Nilpotent Groups and Solvable Groups

Abstract

There are two important properties of groups that are stronger than commutativity: they are solvability and nilpotence. Solvable1 groups are obtained by forming successive extensions of abelian groups; nilpotent groups lie midway between abelian and solvable groups.

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There are two important properties of groups that are stronger than commutativity: they are solvability and nilpotence. Solvable¹ groups are obtained by forming successive extensions of abelian groups; nilpotent groups lie midway between abelian and solvable groups.

5.1 Central Series and Nilpotent Groups

In Chapter 2 we defined the commutator of two elements and the subgroup generated by all the commutators. We now extend these notions.

Definition 5.1.

If x, y and z are three elements of a group G, define [x, y, z] = [[x, y], z], and inductively

(Hence [[x ₁,x ₂, …,x _n − 1],x _n ] = [⋯ [[x ₁,x ₂],x ₃] …, x _n ]). If H, K are subgroups of G, define [H, K] = 〈[h, k], h ∈ H, k ∈ K 〉, and inductively [H ₁, H ₂,…, H _n ] = [[H ₁, H ₂,…, H _n−1], H _n ].

We have [H, K] = [K, H] and [H, K] ⊴ 〈H, K〉. If H, K ⊴ G, then [H, K] ⊴ G and [H, K] ⊆ H ∩ K .If φ is a homomorphism of G, then [H, K]^φ = [H ^φ, K ^φ]. Obviously, [H ₁, H ₂,…, H _n ] ⊇ 〈[h ₁, h ₂,…, h _n ], h _i ∈ H _i 〉, but in general the other inclusion does not hold, as the following example shows.

Example 5.1.

In A ⁵, let H₁ = {I, (1, 2)(3, 4)}, H ₂ = {I, (1, 3)(2, 5)}, H ₃ = {I, (1, 3)(2, 4)}. With h ₁ = (1, 2)(3,4), h ₂ = (1, 3)(2, 5) and h ₃ = (1, 3)(2, 4), we have [h ₁, h ₂, h ₃] = (1, 4, 5, 2, 3), whereas [h ₁, h ₂, h ₃] = 1 if one of the three elements is 1. It follows 〈[h ₁, h ₂, h ₃], h _i ∈ H _i 〉 = 〈(1,4, 5, 2, 3)〉. Now [H ₁, H ₂] contains [h ₁, h ₂]² = (1, 2, 4, 5, 3) and therefore [H ₁, H ₂, H ₃] contains [[h ₁, h ₂]², h ₃] = (1,3,5).

The next lemma follows from the definitions.

Lemma 5.1.

• i) If H, K ≤ G then K ⊆ N _G (H) if and only if [K, H] ⊆ H. In particular, H ⊴ G if and only if [G, H] ⊆ H.

• ii) Let K, H ≤ G with K ⊆ H. Then the following are equivalent:

  • a) K ⊴ G and H/K ⊆ Z(G/K); and

  • b) [H, G] ⊆ K.

Definition 5.2.

A series of subgroups of a group G:

is said to be central if, for all i,

i.e. xH _i , for x ∈ H _i+1, commutes with all the elements of G/H _i . This means that

for all g ∈ G, i.e.

In other words, the action of G on G/H _i given by (xH _i)^g = x ^g H _i is trivial on H _i+1/H _i : (xH _i )^g = xH _i , i.e. x ⁻¹ x ^g ∈ H _i . Conversely, (5.4) implies (5.2). The H _i are normal in G.

We now consider two special kinds of central series. The first one is an ascending series. If in (5.2) we take the entire center Z(G/H _i ) of G/H _i , then setting H _i = Z _i we have:

where Z _i+1/Z _i = Z(G/Z _i ). Hence Z _i+1 consists of all the elements x ∈ G for which (5.3) holds:

Obviously, if the series (5.1) is central, we have H _i ⊆ Z _i .

The second series is descending. Starting from the group G and using (5.4), define the subgroups Γ _i = Γ _i (G) as follows:

(Note that Γ ₂ = [Γ ₁, Γ ₁] = [G, G] = G′, the derived group of G.) We have:

If x ∈ Γ _i then [x, g] ∈ Γ _i+1 for all g ∈ G, so that

i.e. xΓ _i+1 commutes with all the elements of G/Γ _i+1. In other words,

which proves that the series of the Γ _i is central.

By definition of Z _i , if H _n = G then Z _n = G. The next theorem shows that if Z _n = G then Γ _n+1 = {1}.

Theorem 5.1.

Let (5.1) be a central series of a group G with H _n = G. Then:

Proof. The second inclusion has already been seen. As to the first, we prove by induction on j = n − i that Γ _j+1 ⊆ H _n−j ; this inclusion holds for j = 0. Assume it is true for j; then Γ _j+2 = [Γ _j+1, G] ⊆ [H _n–j , G] ⊆ H _n−(j+1), as required. ⋄

Corollary 5.1.

Let

and

Then r = c.

Proof. Let n = c in (5.5). If i = 0, then Γ _c+1 ⊆ Z ₀ = {1}, so c ≥ r. Now let us take for the series of the H _i that of the Γ _i in ascending order:

for i = r we have H _r = Γ ₁ = G and H _r ⊆ Z _r . It follows Z _r = G and r ≥ c. ⋄

In words, this corollary states that in a nilpotent group the series of the Z _i and of the Γ _i reach G and {1}, respectively, in the same number c +1 of steps.

Definition 5.3.

In a nilpotent group G the integer c of Corollary 5.1 is the nilpotence class of G. The series of the Z _i is the upper central series (the ascending central series that reaches G in the least number of steps); the series of the Γ _i is the lower central series (the descending central series that reaches {1} in the least number of steps). It is clear that the terms of both series are characteristic.

The identity group is nilpotent of class c = 0. Abelian groups are nilpotent of class c ≤ 1 (and conversely). If c ≤ 2 then {1} ⊂ Z ⊂ G, so G/Z = Z(G/Z), G/Z is abelian and G′ ⊆ Z; conversely, the latter inclusion implies c ≤ 2. This is the case, for instance, of the group D ₄ or of the quaternion group. The integer c is a measure of how far the group is from being abelian. Note that if G is of class c, G/Z is of class c − 1.

It is clear that if G is nilpotent of class c ≤ m, then [x ₁, x ₂,…, x _m+1] = 1 for all x ₁, x ₂,…, x _m+1 in G, and conversely.

Nilpotence is a property inherited by subgroups and homomorphic images.

Theorem 5.2.

Subgroups and homomorphic images of a class c nilpotent group are nilpotent of class at most c.

Proof.

• i). If H ≤ G, then Γ _i+1(H) = [Γ _i (H), H] ⊆ [Γ _i (G), G] ⊆ Γ _i+1(G), so Γ _i+1(G) = {1}, implies Γ _i+1(H) = {1}.

• ii). If φ is a homomorphism of G, and H, K ≤ G, then [h, k]^φ = [h ^φ,k ^φ], thus Γ _i + 1(G ^φ) = [Γ ^φ _i ,G ^φ] = Γ _i + 1(G)^φ. Hence, if Γ _i+1(G) = {1}, its image equals {1} as well. ⋄

Example 5.2.

The inclusions (5.5) are strict in general. Let G = D ₄ × C ₂ (cf. Ex. 2.11, 3). The center of G is the product of the centers of the factors, so it is a Klein group V, and G/V has order 4 and so is abelian. G is nilpotent with upper central series {1} ⊂ V ⊂ G. The derived group of G coincides with that of D ₄; hence Γ ₂ = C ₂, and this C ₂ being contained in the center of G it follows Γ ₃ = [Γ ₂, G] = {1}. The lower central series is therefore G ⊃ Γ ₂ ⊃ {1}. Hence Γ ₂ is properly contained in Z ₁.

An ideal J of an associative ring R is nilpotent if a power J ⁿ is {0}. By definition,

so J is nilpotent if any product x ₁ x ₂ ⋅ ⋅ ⋅ x _n of n elements is zero. We now show that in a ring with unity a nilpotent ideal gives rise to a nilpotent group (a subgroup of the invertible elements of the ring).

Theorem 5.3.

Let J be a nilpotent ideal of a ring R with unity 1, J ⁿ = {0}. Then G =1 + J = {1 + x, x ∈ J} is a nilpotent group.

We prove that G satisfies the three properties of a group.

• i). closure: (1 + x)(1 + y) = 1 + (x + y + xy), and xy ∈ J if x, y ∈ J;

• ii). identity: 1;

• iii). inverse: (1 + x)(1 − x + x ² − ⋅ ⋅ ⋅ ± x ⁿ⁻¹) = 1.

Moreover, G is nilpotent. Let H _k = 1 + J ^k. The ideal J ^k is nilpotent because (J ^k)ⁿ = J ^nk = {0}, so by 1 the H _k are subgroups. Moreover, J ^k = J ^k-1 ⋅ J ⊆ J ^k−1, thus H _k ⊆ H _k−1. Finally, the series

is central. Let us show that

i.e. [a,g] ∈ 1 + J ^k + 1 or [a,g] − 1 ∈ J ^k + 1. Now,

and with a = 1 + x, x ∈ J ^k, and g = 1 + y, y ∈ J, we have

so [a, g] − 1 ∈ J ^k+1. ⋄

Examples 5.3.

1. In the ring of n × n upper triangular matrices, those having zero diagonal form a nilpotent ideal J with J ⁿ =0. If I is the identity matrix, G = I + J is the group of upper unitriangular matrices. Over a finite field, G is a nilpotent group of order q ^n(n−1)/2 (see Ex. 3.4, 1).

2. If V is a vector space of dimension n over a field K and

where V = 〈υ ₁,υ ₂,…,υ _n 〉 and V _n−i = 〈υ _i+1,…, υ _n 〉, the linear transformations ϕ such that V _i ϕ ⊆ V _i − 1 form an ideal J which is nilpotent because

for all υ ∈ V, so ϕ ₁ ϕ ₂ … ϕ _n =0 and J ⁿ = {0}. The matrices of the linear transformations ϕ are the matrices of the ideal J of the previous example. The group G of Theorem 5.3 stabilizes the series of subspaces V _i (we recall that this means that if υ ∈ V _i and g ∈ G, then υ ^g ∈ V _i and υ ^g − υ ∈ V _i−1, i.e. G fixes the subspaces and acts trivially on the quotient spaces).

Let us consider a few properties of nilpotent groups.

Theorem 5.4.

If G is nilpotent and H< G, then H<N _G (H)) (“normalizers grow”).

Proof. Let {Г _k } be the lower central series, and let i be such that Г _i ⊈ H and Г _i+1 ⊆ H. Then [Г _i , H] ⊆ [Г _i , G] = Г _{i +1} ⊆ H, so Г _i normalizes H and is not contained in H. ⋄

Corollary 5.2.

A maximal subgroup of a nilpotent group is normal, and therefore is of prime index.

Corollary 5.3.

In a nilpotent group, the derived group is contained in the Frattini subgroup.

Proof. A maximal subgroup M is normal, hence G/M is of prime order and G′ ⊆ M. This holds for all maximal subgroups, so G′ ⊆ ∩M = Φ(G). ◊

Theorem 5.5.

Let G be nilpotent, N ≠ {1} a normal subgroup. Then the intersection N ∩ Z(G) = {1}. In particular, every minimal normal subgroup of G is contained in the center and has prime order.

Proof. Let { Г _k } be the lower central series of G and let i be such that Г _i ∩N ≠ {1} and Г _{i +1} ∩N = {1}. Then [Г _i ∩N, G] ⊆ [Г _i , G] ⊆ Г _{i +1} because {Г _k } is central, and [Г _i ∩ N, G] ⊆ N because N is normal. It follows [Г _i ∩ N, G] ⊆ Г _{i +1} ∩ N = {1}, and therefore N ∩ Z(G) ≠ {1} because it contains Г _i ∩ N = {1}. If N is minimal, then N ⊆ Z(G): the subgroups of N are normal in G, so N has no proper subgroups and |N| = p. ◊

These properties have already been encountered in the case of finite p- groups. The reason is in the following theorem.

Theorem 5.6.

A finite p-group of order p ⁿ is nilpotent, and is of class at most n − 1.

Proof. Since the center of a p-group is nontrivial, the upper central series stops when G/Z _i = 1, i.e. Z _i = G. Hence the group is nilpotent. If it is of class c, then:

All the quotients have order at least p because they are all nontrivial, but not all of them have order p. Indeed, Z _c /Z _c−1= G/Z _c−1 cannot be cyclic; if it were, the quotient of G/Z _c−2 by the center Z _c−1/Z _c−2 would be cyclic because it is isomorphic to G/Z _c−1, and G/Z _c−2 would be abelian (Chapter 2, ex. 23) and would coincide with its center Z _c−1/Z _c−2. Then G = Z _c−1, against G being of class c. There being c quotients, the product of their orders is greater than p ^c, so p ⁿ > p ^c, c < n and c < n ₋1. ◊

A p-group of order p ⁿ always admits a central series of length n + 1 (ex. 2), and therefore with cyclic quotients of order p. For example, in D ₄ we have the series {1} = Z ₀ ⊂ Z ₁ ⊂ V ⊂ D ₄, with V in between Z ₁ and Z ₂ = D ₄.

An infinite p-group is not necessarily nilpotent (ex. 5). The next theorem shows that in the finite case nilpotent groups are a generalization of p-groups.

Theorem 5.7.

In a finite nilpotent group a Sylow p-subgroup is normal, and therefore is the unique Sylow p-subgroup. It follows that the group is the direct product of its Sylow p-subgroups, for the various primes p dividing the order of the group.

Proof. If S is a Sylow p-subgroup, then by Theorem 3.16 N _G (N _G (S)) = N _G (S). But by Theorem 5.4, if N _G (S) ≠ G then N _G (N _G (S)) > N _G (S). Hence N _G (S) = G, and S is normal. The product of these p-Sylows for the various p is then a subgroup that coincides with G because its order is divisible by the whole power of p dividing |G|, for all p. An element x _i of a p _j -Sylow S _i cannot equal a product of elements x _j of p _j -Sylow S _j , j ≠ i, because since the p _i -Sylows are pairwise commuting (they are normal and of trivial intersection) this product has order the product of the orders (the x _j have coprime orders), and therefore is coprime to o(x _i ). Hence a Sylow p-subgroup has trivial intersection with the product of the others, and therefore the product of the Sylow p-subgroups is direct. ◊

Theorem 5.8.

The Frattini subgroup Φ = Φ(G) of a finite group G is nilpotent.

Proof. Let P be a p-Sylow of Φ, and let us show that P is normal in Φ (in fact P ⊴ G). Since Φ ⊴ G, by the Frattini argument (Theorem 3.15) G = ΦN _G (P). A fortiori, G = 〈Φ, N _G (P)〉, so by the property of Φ, G = 〈(N _G (P)) =N _G(P)), i.e. P ⊴ G. It follows that the group is the direct product of its Sylows and so (ex. 3) is nilpotent. ◊

It is not true in general that if N is a nilpotent normal subgroup of a group G such that G/N is nilpotent then G is nilpotent²; the group S ³ is a counterexample. However, if N is the center or (in the finite case) the Frattini subgroup then G is nilpotent (see also ex. 20, iii)).

Theorem 5.9.

• i). If G/Z(G) is nilpotent, then so is G;

• ii). if G is finite and G/Φ(G) is nilpotent, then so is G.

Proof.

• 1) Let Z = Z(G). From Γ _i+1(G/Z) = Γ _i Z/Z and Γ _n+1(G/Z) = Z it follows Γ _n Z/Z = Z, from which Γ _n Z ⊆ Z, Γ _n ⊆ Z and Γ _n+1 = {1}.

  In the finite case, the proof can go as follows. Let Z = Z(G), with G/Z nilpotent, and let SZ/Z be a p-Sylow of G/Z, some p. Since SZ/Z ⊴ G/Z we have SZ ⊴ G. Then, by the Frattini argument, G = Z N _G (S), and since Z ⊆ N _G (S), we have G = N _G (S) and S ⊴ G. But this holds for all p, so G is nilpotent.

• ii Let Φ = Φ (G), and let G / Φ (G) be nilpotent. As above, G = Φ N _G (S). Hence G = 〈Φ, N _G (S)) =(N _G(S)〉 = N _G (S), and S ⊴ G. ◊

Lemma 5.2.

• i) Let x, y, z ∈ G; then:

  

  (Hall-Witt identity).

• ii) (Hall’s three subgroups lemma) Let H, K, L ≤ G and N ⊴ G. If [H, K, L] ⊆ N and [K, L, H] ⊆ N, then [L, H, K] ⊆ N.

• iii) $\begin{array}{c}\hfill \left[xy,z\right]={\left[x,z\right]}^y\left[y,z\right]=\left[x,z\right]\left[\left[x,z\right],y\right]\left[y,z\right]\hfill \\ \hfill \left[x,yz\right]=\left[x,z\right]{\left[x,y\right]}^z=\left[x,z\right]\left[x,y\right]\left[\left[x,y\right]z\right].\hfill \end{array}$

Proof.

• i) We have

  

  and therefore [x, y ⁻¹, z] = x ⁻¹ y ⁻¹ xz ⁻¹ · yxy ⁻¹ zy = a · b. From i) we have, similarly,

  

• ii) For x ∈ H, y ∈ K and z ∈ L we have, by assumption,

  

  Then, by i), is also [z, x ⁻¹, y]^x ∈ N and therefore [z, x ⁻¹, y] ∈ N for all z ∈ L, x ∈ H, y ∈ K, and so also [L, H, K] ⊆ N.

• iii) follows by calculation. ◊

The linear transformations of a vector space that stabilize the series of subspaces given in Ex. 5.3, 2 form a nilpotent group. The next theorem shows that in the fact of stabilizing a series of subgroups lies the true reason for nilpo-tence. We prove the theorem in the case of an invariant series of subgroups, but by a result of Hall it also holds without the hypothesis of normality on the subgroups³ (in this case, however, the upper bound for the nilpotence class is far worse: if the series has n +1 terms, one obtains a nilpotent group of class at most $\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2\hfill \end{array}\right)$ ).

Theorem 5.10.

Let G = G ₀ ⊇ G ₁ ⊇…⊇ G _n = {1} be an invariant series ⁴ of G having n +1 terms, and let A be the group of automorphisms of G that stabilize the series, that is,

For α ∈ A and i = 0, 1,…, n − 1. Then A is nilpotent of class less than n.

Proof. Consider the subgroups A _j of A defined as follows:

Therefore A _j consists of the elements of A acting trivially on the quotients G _i /G _i+j , i = 0, 1,…,n − j. Clearly,

Let us prove that (5.7) is a central series, i.e.

for all j. Consider the semidirect product $\overline{G}$ of G by A. Then an element of the form x ⁻¹ x ^α is a commutator of $\overline{G}$, and the second condition of (5.6) means that [G _i , A] ⊆ G _i+i . Then (5.8) is equivalent to

because, by definition, A _j+1 consists of all elements α such that [G _i , α] ⊆ G _i+j+1. But

and

and being G _i+j+1 ⊴ G, by Lemma 5.2, ii), we have (5.9). Since (5.7) is of length at most n the result follows. ◊

Lemma 5.3.

If H, K and L are normal in a group G, then [HK, L] = [H, L] [K, L] and [H, KL] = [H, K][H, L].

Proof. For x ∈ H, y ∈ K and z ∈ L we have [xy, z] = [x, z]^y[y, z] = [x ^y, z ^y][y, z] (Lemma 5.2, iii). H and L being normal, [x ^y, z ^y] ∈ [H, L], from which [HK, L] ⊆ [H, L][K, L]. The other inclusion is obtained by observing that [H, L], [K, L] ⊆ [HK, L]. Similarly for the second equality. ◊

Theorem 5.11 (Fitting).

The product of two normal nilpotent subgroups H and K of a group G, of class c ₁ and c ₂, respectively, is a normal nilpotent subgroup of class at most c ₁ + c ₂.

Proof. We may assume G = HK. We have

Repeated application of Lemma 5.3 yields Γ _n (G) as a product of 2ⁿ terms each of which has the form A = [A ₁, A ₂,…, A _n ], where the A _i are equal to either H or K. H and K are normal in G, and Γ _i (H) is characteristic in H and therefore normal in G, from which [Γ _i (H), K] ⊆ H; similarly for K. It follows that if in A there are l of the A _i that are equal to H, then A ⊆ Γ _l (H); similarly A ⊆ Γ _n−l (K), from which A ⊆ Γ _l (H) ∩ Γ _n−l (K). With n = c ₁ + c ₂ + 1, we have either l ≥ c ₁ + 1 or n − l ≥ c ₂ + 1. In any case A = {1}. ◊

Let us prove directly that under the hypothesis of the previous theorem the center of G is nontrivial. If H ∩ K = {1}, then H and K commute elementwise. So {1} ≠ Z(H) centralizes K, and since it also centralizes H, it centralizes G. Hence Z(G) contains Z(H) that is different from {1}. Thus assume H ∩ K ≠ {1}. H ∩ K is a nontrivial normal subgroup of H, so it meets the center of H non trivially (Theorem 5.5): Z(H) ∩ H ∩ K ≠ {1}. It follows, Z(H) ∩ K ≠ {1}, and since Z(H) ⊴ G (it is characteristic in H ⊴ G), we have that Z(H) ∩ K is normal in G, and in particular in K. Hence Z(K) ∩ Z(H) ∩ K ≠ {1}, so Z(K) ∩ Z(H) ≠ {1}. The elements of this intersection centralize H and K, and therefore G. It follows that Z(G) ≠ {1}.

In the finite case it is possible to deduce from this proof that G is nilpotent. Indeed, set Z = Z(G); then G/Z is the product of the normal nilpotent subgroups HZ/Z and KZ/Z. But |G/Z| < |G|, so by induction G/Z is nilpotent, and so also is G (Theorem 5.9, i)).

The next theorem characterizes in various ways the nilpotent finite groups.

Theorem 5.12.

The following properties of a finite group G are equivalent:

• i). G is nilpotent;

• ii). if H < G then H<N _G (H);

• iii). maximal subgroups are normal;

• iv). G′ ⊆ Φ(G);

• v). Sylow p-subgroups are normal;

• vi). G is the direct product of its Sylow subgroups;

• vii). G is the product of normal p-subgroups, for various p.

Proof.The implications

• i) ⇒ ii) ⇒ iii) ⇒ iv) are those of Theorem 5.4 and its Corollaries 5.14 and 5.15;

• iv) ⇒ v) (Wielandt) SG′ ⊴ G, so by the Frattini argument it follows G = SG′ N _G (S) = 〈S, G′, N _G (S)〉 = 〈G′, N _G (S)〉 = N _G (S), where the last equality follows from the assumption G′ ⊆ Φ(G);

• v) ⇒ Theorem 5.7;

• vi) ⇒ obvious;

• vii) ⇒ Theorem 5.6 and Theorem 5.11. ◊

Definition 5.4.

The maximal nilpotent normal subgroup of a group G is called the Fitting subgroup of G. If G is finite such a subgroup always exists, and is the product of all normal nilpotent subgroups of G. It is denoted F(G). (It is not excluded that F(G) = {1}, as is the case of a simple group.)

In the finite case, the Fitting subgroup has the following characterization (recall that O _p (G) is the maximal normal p-subgroup of G ⁵).

Theorem 5.13.

Let G be a finite group, and let p _i , i = 1, 2, …, t, be the prime divisors of the order of G. Then:

Proof. Since O _p ⊆ F(G) for all p we have one inclusion. As to the other, observe that F(G) being nilpotent, it has a normal Sylow p-subgroup, which as such is contained in all the Sylow p-subgroups of G, and so in O _p . But F(G) is the product of its Sylow p-subgroups, and the inclusion follows. ◊

Theorem 5.14.

Let G be a finite group, and let F = F(G), Φ = Φ (G), Z = Z(G). Then:

• i). Φ ⊆ F, Z ⊆ F;

• ii). F/Φ = F(G/Φ);

• iii). F/Z = F(G/Z).

Proof. i) is obvious. As for ii) we have, for one thing, F/Φ ⊆ F (G/ Φ) because F/Φ is the image of the nilpotent normal subgroup F under the canonical homomorphism. As to the other inclusion, observe that if H/Φ is the Fitting subgroup of G/Φ, and P is a Sylow p-subgroup of H, then P Φ/Φ is Sylow in H/Φ, and therefore characteristic, by the nilpotence of H/Φ; hence it is normal in G/Φ. It follows P Φ ⊴ G. Now apply the Frattini argument as in Theorem 5.8. iii) is proved similarly. ◊

Theorem 5.15.

The Fitting subgroup of a finite group G centralizes all the minimal normal subgroups of G.

Proof. Let N be minimal normal in G, and let H be normal and nilpotent. If N ∩ H = {1}, then N and H commute elementwise, so H centralizes N. If N ∩ H ≠{1}, then N ⊆ H by the minimality of N. H being nilpotent, N ∩ Z(H) = {l}. But Z(H) characteristic in H and H ⊴ G imply Z(H) ⊴ G, and always because of the minimality of N, N ⊴ Z(H); hence H centralizes N in this case too. ◊

Corollary 5.4.

Let H/K be a chief factor of a group G. Then F(G/K) ⊆ C _G/K (H/K).

An element gK centralizes the quotient H/K if [h, g] ∈ K for all h ∈ H.

Definition 5.5.

If K ⊆ H are two normal subgroups of a group G, the centralizer in G of the factor H/K is the subgroup

Hence, by definition, C _G/K (H/K) = C _G (H/K)/K; the quotient H/K being normal in G/K its centralizer C _G/K (H/K) is also normal. The inclusion of Corollary 5.4 may therefore be written as F(G/K) ⊆C _G (H/K). F K/K is normal in G/K and nilpotent, so is contained in F(G/K), and therefore F(G) ⊆ C _G (H/K). Summing up:

Corollary 5.5.

The Fitting subgroup of a finite group G centralizes all minimal normal subgroups of a quotient of G. In particular, it centralizes every chief factor of G.

Chief series and nilpotent normal subgroups are related by the following theorem.

Theorem 5.16.

Let {G _i } be a chief series of a finite group G, i = 1, 2,…, n. Then:

Proof. Set L = ∩  ^n − 1 _i = 0 C _G (G _i /G _i + 1). By Corollary 5.5 we have the inclusion F(G) ⊆ L. As to the other inclusion, let us prove that L is normal and nilpotent. Obviously, it is normal. Let us consider the series {L _i }, i = 0, 1,…, n−1, where L _i = L ∩ G _i , and let us prove that it is a central series of L. By definition of C _G (G _i/G _i+1), we have [G _i , C _G (G _i /G _i +₁)] ⊆ G _i +₁, and since L _i ⊆ G _i , it follows:

which, together with [L _i , L] ⊆ L, yields the inclusion [L _i, L] ⊆ L _i+1, showing that the series of the L _i is central. Hence, L is a nilpotent normal subgroup of G, so L ⊆ F(G). ◊

We now consider a few properties of nilpotent groups in the general case.

Theorem 5.17.

A f.g. periodic nilpotent group G is finite.

Proof. By induction on the nilpotence class c of the group. If c = 1, then G is abelian, f.g. and periodic, and therefore finite. Let c>1. G/Z _c−1 is f.g., periodic and abelian, and therefore finite; Z _c−1 being of finite index is f.g., and being periodic and of class c − 1 by induction is finite, and so is G. ◊

Theorem 5.18.

A f.g. nilpotent group with finite center is finite.

Proof. Fix g ∈ G and consider the mapping from Z ₂ to the center of G given by x → [x, g]. This mapping is a homomorphism:

because [x, y] ∈ Z and therefore [x, g]^y = [x, g]. If G = 〈g ₁, g ₂,…, g _n 〉, there are n homomorphisms Z₂ → Z given by x → [x, g _i ], with kernels Z ₂ ∩ C _G (g _i ), each of which has finite index (the images are contained in the center, which is finite). Their intersection is Z ₂ ∩ (∩ ⁿ _i = 1 C _G (g _i )) = Z ₂ ∩ Z = Z, so Z has finite index in Z ₂, and therefore Z ₂ is finite. Now Z(G/Z) = Z ₂/Z, which is finite, and G/Z has class c − 1 and is f.g. By induction on c, G/Z is finite and so also is G (if c =1, then G = Z, and there is nothing to prove).◊

Let now G be a group generated by a set X. The group G/G′ = Γ ₁/Γ ₂ is generated by the images xΓ ₂, x ∈ X, and since Γ ₂ is generated by the commutators [x, y], x, y ∈ G, the quotient group Γ ₂/Γ ₃ is generated by the images of these commutators. Let us show that the commutators between elements of X suffice, i.e. that

Indeed,

which, by the first equality of Lemma 5.2, iii), and setting x = x _i , y = g, z = x _j h, equals

where the last two terms belong to Γ ₃. As to the first term we have, by the second equality of Lemma 5.2, iii),

where the first and third term belong to Γ ₃. It follows [x, y] ∈ 〈[x _i , x _j ], Γ ₃〉 [x,y] ∈ 〈[x _i ,x _j ],Γ ₃〉 and (5.10). In general, let us prove by induction that

(recall that[x ₁, x ₂, …, x _i ] means [[… [x ₁, x ₂], x ₃ ],…, x _i ]), which we call a commutator of weight i. Assume (5.11) holds. Now

and by (5.11) x = t ₁ t ₂ … t _k s, where the t _k are commutators of weight i in the generators in X and s ∈ Γ _i+1. It follows

with the last two terms in Γ _i+2. If we now write y in terms of the generators in X we have:

where the last term of the right hand side is in Γ _i+2. As to the first term observe that

and similarly for the second. Proceeding in this way we obtain:

and the following result:

Theorem 5.19.

If a group is generated by a set X, then the i-th term of the lower central series Γ _i is generated by the commutators of weight i in the elements of X and by the elements of Γ _i+1.

Corollary 5.6.

If a group is f.g, the quotients of the lower central series are also f.g.

Corollary 5.7.

A f.g. nilpotent group admits a central series with cyclic quotients.

Proof. The quotients of the lower central series are f.g. and abelian, and as such are direct product of cyclic group. For i = 1, 2,…, c, let:

with the A _k/Γ _i+1 cyclic. Then, setting G _k /Γ _i+1 = A _k /Γ _i+1 ×…× A _n /Γ _i+1, we have (Theorem 2.30, ii)):

a cyclic group. Hence the quotients of the series Γ _k = G ₁ ⊃ G ₂ ⊃ … ⊃ G _n = A _n ⊃ Γ _i+1 are cyclic. ◊

By virtue of the last corollary, we have that a f.g. nilpotent group is built up by finitely many successive extensions of cyclic groups; we call cyclic an extensions H of a group K with H/K cyclic. We have the following theorem:

Theorem 5.20.

A cyclic extension of a f.g. and residually finite group G is residually finite.

Proof. Let G = 〈a, H〉, with H ⊴ G residually finite, and let 1 ≠ g ∈ G. If g ∉ H, G/H being cyclic, and therefore residually finite, there exists L/H ⊲ G/H such that (G/H)/(L/H) is finite and gH ∉ L/H; then g ∉ L, with G/L finite. Therefore we may assume g ∈ H. Let g ∉ K ⊲ H and H/K finite. H being f.g., there exists a characteristic subgroup of H, still denoted K, of finite index in H (Theorem 3.5, ii)) not containing g (there are only a finite number of subgroups of a given finite index (Theorem 3.5, i))). Moreover, K characteristic in H and H ⊲ G imply K ⊲ G. If G/H is finite, G/K is also finite, and g ∉ K. Thus we assume G/H = 〈aH〉 infinite, so that 〈aH〉 = 〈a〉H with 〈a〉 ∩ H ={1}. By the N/C theorem, since H/K is normal in G/K and finite, the centralizer C _G/K (H/K) has finite index in G/K. It follows that a power of aK centralizes H/K, (aK)^m ∈ C _G/K (H/K), m ≠ 0, and this (aK)^m has infinite order because if a ^mt ∈ K, then a ^mt ∈ H, whereas 〈aH〉 is infinite cyclic. It follows gK ∉ 〈a ^m K〉 (gK belongs to the finite group H/K). Now a ^m K commutes with all the elements hK, and obviously with aK, and therefore with all the elements of G/K; in particular (a ^m K) ⊲ G/K. The quotient (G/K)/(〈a ^m K〉/K) is finite, and the image of g in this group is not the identity. ◊

Definition 5.6.

A group is polycyclic if it admits a normal series with cyclic quotients.

Theorem 5.21.

A polycyclic group is f.g., and so are its subgroup.

Proof. Let G = G ₀ ⊃ G ₁ ⊃ … ⊃ G _i ⊃ … ⊃ G _n = {1} be a normal series of G with cyclic quotients. We have G/G ₁ = (xG ₁ for a certain x∈G, and therefore G = 〈x, G ₁〉 . By induction on n, G ₁ is f.g., and therefore G is. Similarly, all the other G _i of the series are f.g. If H ≤ G, then HG ₁/G ₁ ≃ H/(H ∩ G ₁), and therefore H/(H ∩ G ₁) is cyclic because is isomorphic to a subgroup of the cyclic group G/G ₁. Then H = 〈x, H ∩ G ₁〉 for some x. But by induction H ∩ G ₁ is f.g. because is a subgroup of G ₁ which is f.g. and polycyclic, with a normal series with cyclic quotients of length n − 1. It follows that H is f.g. ◊

As in the abelian case we have (cf. Theorem 4.2):

Corollary 5.8.

In a f.g. nilpotent group every subgroup is f.g.

Corollary 5.9.

A f.g. nilpotent group is residually finite, and therefore hopfian.

Proof. By Theorem 5.20 a polycyclic group is residually finite, and since by Corollary 5.7 a f.g. nilpotent group is polycyclic, by Theorem 4.39 we have the result. ◊

Corollary 5.10.

For a f.g. nilpotent group, and more generally for polycyclic groups, the word problem is solvable.

Proof. Recall that if N ⊴ G and G/N are finitely presented, then so also is G (Chapter 4, ex. 46). Since a polycyclic group is obtained by a sequence of extensions of finitely presented groups by means of cyclic groups, a polycyclic group is finitely presented; it being residually finite by Corollary 5.9, the result follows from Theorem 4.38. ◊

Lemma 5.4.

If G is nilpotent of class c ≥ 2 and x ∈ G, then H = (x, G′) is of class at most c−1.

Proof. Let us show that H = Z _c−1(H). We have G′ ⊆ Z _c−1(G)∩H ⊆ Z _c−1(H), so H/Z _c−1(H) ≃ (H/G′)/(Z _c−1(H)/G′) is a cyclic group, being a quotient of the cyclic group H/G′. But

and therefore H/Z _c−2(H) is abelian (the quotient w.r.t. the center is cyclic). Hence, the quotient appearing on the right hand side of the previous expression is the identity, so that H = Z _c−1(H).

(Note that this lemma holds not only for G′ but also for every normal subgroup of G contained in Z _c−1.)◊

Theorem 5.22.

In a torsion-free nilpotent group we have:

• i). an element x ≠ 1 cannot be conjugate to its inverse;

• ii). if x ⁿ = y ⁿ then x = y;

• iii). if x ^h y ^k = y ^k x ^h then xy = yx.

Proof.

• i). Let us show that if x ∼ x ⁻¹, either x = 1, or else the order of x is power of 2. By induction on the nilpotence class c. If c = 1, then G is abelian, and x ∼ x ⁻¹ means x = x ⁻¹, i.e. x ² = 1. Let c > 1. If x belongs to the center Z we still have x ² = 1; if x ∉ Z, then xZ ∼ x ⁻¹ Z and by induction (xZ)^2k = Z, i.e. ${x}^{2^k}\in Z.$ But since ${x}^{2^k}\sim {x}^{-{2}^k}$, we have ${x}^{2^k}={x}^{-{2}^k}$ and ${x}^{2^{k+1}}=1.$.

• ii). By Lemma 5.4, the subgroup H = 〈x, G′〉 is nilpotent of class at most c − 1. Now, x ⁻¹ y ⁻¹ xy ∈ G′, and therefore x · x ⁻¹ · y ⁻¹ xy = y ⁻¹ xy ∈H, so (y ⁻¹ xy)ⁿ = x ⁿ since (y ⁻¹ xy)ⁿ = y ⁻¹ x ⁿ y = y ⁻¹ y ⁿ y = y ⁿ = x ⁿ. By induction on c we have y ⁻¹ xy = x, i.e. x and y commute; from x ⁿ = y ⁿ then follows (xy ⁻¹)ⁿ = 1, and torsion-freeness implies xy ⁻¹ = 1 and x = y.

• iii). We have y ^−k x ^h y ^k = x ^h i.e. (y ^−k xy ^k)^h = x ^h. By ii), it follows y ^−k xy ^k = x, from which x ⁻¹ y ^k x = y ^k, i.e. (x ⁻¹ yx)^k = y ^k, and again by ii) x ⁻¹ yx = y, i.e. xy = yx. ◊

Remark 5.1.

The properties of Theorem 5.22 are shared by free groups.

Theorem 5.23.

In a torsion-free nilpotent group G the quotients of the upper central series are torsion-free.

Proof. Consider Z ₂/Z ₁, and let (xZ ₁)^h = Z ₁ for some x ∈ Z ₂. Then x ^h ∈ Z ₁ and therefore x ^h y = yx ^h, all y ∈ G. By iii) of Theorem 5.22, xy = yx, all y ∈ G, i.e. x ∈ Z ₁. Similarly, if x ∈ Z ₃ and x ^h G Z ₂ then, by definition, [x, y] ∈ Z ₂ and [x ^h, y] ∈ Z ₁, all y ∈ G, i.e. x ^−h y ⁻¹ x ^h y = x ^−h(y ⁻¹ xy)^h ∈ Z ₁. But then x ^−h and (y ⁻¹ xy)^h commute, so x ⁻¹ and y ⁻¹ xy also commute. Hence x ^−h(y ⁻¹ xy)^h = [x, y]^h = (x ⁻¹ y ⁻¹ xy)^h ∈ Z ₁, and being [x, y] ∈ Z ₂, by the previous case we have [x, y] ∈ Z and x ∈ Z ₂. This holds in general: if Z _i /Z _i−1 is torsion-free Z _i+1/Z _i is also torsion-free. We must show that if x ∈ Z _i +₁, i.e. [x, y] ∈ Z _i , all y ∈ G, and x ^h ∈ Z _i , i.e. [x ^h, y] ∈ Z _i−1, then x ∈ Z _i , i.e. [x, y] ∈ Z _i−1. The quotient Z _i /Z _i−1 being torsion-free, it suffices to show that [x, y]^h ∈ Z _i−1, i.e.:

Since x ^h ∈ Z _i , we have y ⁻¹ x ^−h y = (y ⁻¹ x ⁻¹ y)^h ∈ Z _i and also (y ⁻¹ x ⁻¹ y)^h x ^h ∈ Z _i , and since Z _i /Z _i−1 is the center of G/Z _i−1, the two cosets (y ⁻¹ x ⁻¹ y)^h Z _i−1 and x ^h Z _i−1 commute. But Z _i /Z _i−1 is torsion-free, so y ⁻¹ x ⁻¹ yZ _i−1 and xZ _i−1 also commute; it follows:

But y ⁻¹ x ^−h yx ^h = [y, x ^h] = [x ^h, y]⁻¹ ∈ Z _i−1, and so also [y, x]^h ∈ Z _i−1 and [x, y]^h ∈ Z _i−1. ◊

Theorem 5.23 is not true in general for the quotients of the lower central series Γ _i /Γ _i+1, as the following example shows.

Example 5.4.

The Heisenberg group ℋ (Ex. 4.3) is torsion free, and is nilpotent of class 2 because the factor group w.r.t. the center is abelian. Consider in ℋ the subgroup of the triples having in the first position a multiple of a fixed integer n:

For the commutator of two elements of G we have:

and on the other hand (0, 0, n) = [(2n, 1, c), (n, 1, c′)]. It follows:

The element (0,0,1) of G is such that (0, 0,1)ⁿ = (0,0, n) ∈ Γ ₂, and therefore in Γ _i /Γ ₂ the coset to which (0, 0,1) belongs has finite period.

We close this section with one more application of Lemma 5.4, which shows a property of nilpotent groups shared by abelian groups.

Theorem 5.24.

The elements of finite period of a nilpotent group G form a subgroup.

Proof. By induction on the nilpotence class c of the group G. If c=1, the group is abelian. Let c >1 and let x and y be two elements of finite period. We must show that xy also has finite period. Let H = 〈x, G′〉; by Lemma 5.4, H is of class less than c, so by induction its elements of finite period form a subgroup t(H). This is characteristic in H, which is normal in G (it contains G′) and therefore is normal in G. Similarly, if K = 〈y, G′〉, then t(K) ⊴ G. Let m = o(x); then:

with y′ ∈ t(K). If o(y′) = n then (xy)^mn = 1, as required. ◊

5.2 Exercises

1. The subgroups Z _i and Γ _i are characteristic in any group.

2. A group of order pn admits a central series of length n + 1 in which all the quotients have order p. [Hint: with H central of order p, the group G/H admits by induction a central series of length n.]

3. The direct product of a finite number of nilpotent groups is nilpotent.

4.

• i Prove by induction on m, that for p prime, ${\left(1+p\right)}^{p^{m-1}}\equiv 1\mod {p}^m$; [Hint: for 0 < i < p, $\left(\begin{array}{c}\hfill p\hfill \\ \hfill i\hfill \end{array}\right)$ is divisible by p.]

• ii) prove that in the cyclic group 〈a〉 of order p ^m the mapping b : a → a ^1+p is an automorphism of order p ^m−1;

• iii) prove that the semidirect product of 〈a〉by 〈b〉of ii) is a nilpotent p-group of class n. [Hint: prove that $G\supset \langle {a}^p\rangle \supset \langle {a}^{p^2}\rangle \supset \dots \supset \langle {a}^{p^{m-1}}\rangle \supset \left\{1\right\}$ is a central series of G.] Hence, for each n, we have a nilpotent group of class n, and in fact a p-group.

5. Let p be a prime, and let G = P ₁ × P ₂ ×…× P _n ×… be an infinite direct product of p-groups P _n of class n, n = 1, 2,.... Prove that:

• i) G is not nilpotent; (Hence an infinite p-group is not necessarily nilpotent; moreover, ex. 3 is not true for an infinite number of groups.)

• ii) the union of a chain of normal nilpotent subgroups is normal but not necessarily nilpotent;

• iii) G has no maximal normal nilpotent subgroups.

6. If G is nilpotent, and H < G, then HG′ < G. [Hint: let HZ _i+1 = G and HZi < G; then HZ _i ⊴ HZ _i+i with abelian quotient, so that G′ ⊆ HZ _i < G.]

7. If x ∈ Z ₂, then x commutes with all its conjugates.

8. In a finite p-group of order p ⁿ and class n − 1 we have Z _i ⊆ Γ _n−i .

9. In a finite nilpotent group, a normal subgroup of order p ^m, p a prime, is contained in the m-th term of the upper central series. [Hint: Theorem 5.5.]

10. A finite group is nilpotent if and only if any two elements of coprime order commute. [Hint: the centralizer of a p-Sylow contains all the q-Sylows, q ≠ p.]

11. A dihedral group D _n is nilpotent if and only if n is a power of 2.

12. If G is a nonabelian nilpotent group, and A is a maximal abelian normal subgroup, then C _G (A) = A.

13. G/F(G) nilpotent does not imply G nilpotent.

14. Let G be a finite nilpotent subgroup (not cyclic). Then:

• i) there exist in G two subgroups of the same order that are not conjugate;

• ii) G cannot be generated by conjugate elements.

15. Let M be a nilpotent maximal subgroup of a finite group G. Prove that either M is of order coprime to its index, or for some p a p-Sylow of M is normal in G.

16. Let H be a maximal nilpotent subgroup of a finite group G. Prove that N _G (N _G (H)) = N _G (H).

17. Let G be a finite group, N ⊴ G, and H minimal among the subgroups whose product with N is G. Prove that:

• i) N∩H ⊆ Φ(H);

• ii) if G/N is nilpotent (cyclic), then H is nilpotent (cyclic).

18. Let C be a class of finite groups such that:

• i) if G/Φ(G) ∈C then G ∈C;

• ii) if G ∈ C and N ⊴ G then G/N ∈ C.

Prove that if N ⊴ G and G/N ∈C, then there exists H ≤ G with H ∈C such that G = NH. [Hint: let H be minimal such that G = NH; prove that N ∩ H ⊆ Φ(H).] (Examples of classes with the given properties are the class of nilpotent groups and that of cyclic groups.)

19. Let A, B, C be three subgroups of a group G such that A ⊆ C ⊆ AB. Prove that C = AB∩ C = A(B ∩ C) (Dedekind’s identity). [Hint: consider c = ab.]

20.

• i) Let N ⊴ G. Prove that Φ(N) ⊆ Φ(G). [Hint: use Dedekind’s identity.]

• ii) Give an example to show that that in i) the normality assumption is necessary. [Hint: let G be the semidirect product of C ₅ = 〈a〉 by C ₄ = 〈b〉, where b ⁻¹ ab = a ².]

• iii) If N ⊴ G and G/N′ are nilpotent, then G is nilpotent. [Hint: use i).]

21. A finite group such that for each prime p there exists a composition series a term of which is a p-Sylow is nilpotent.

22. Let G be a finite nonabelian group with non trivial center and such that every proper quotient is abelian. Prove that G is a p-group.

23. Let H ≤ G, and let A be the group of automorphisms of G such that x ⁻¹ x ^α ∈ H for all x ∈ G and h ^α = h for all h ∈ H (this is the case n = 2 of Theorem 5.10 with G ₁ not necessarily normal). Prove that A is abelian.

24. The following are equivalent:

• i) [[a, b], b] = 1, for all a, b ∈ G;

• ii) two conjugate elements commute.

25. If H is a subnormal subgroup of a group G (Definition 2.16) and N a minimal normal subgroup of G, then N normalizes H. [Hint: let H ⊲ H ₁ ⊲ H ₂ ⊲ … ⊲ H _n ⊲ G; by induction on n. If n = 1, H ⊴ G; if n> 1 and N ∩ H _n = {1}, N ⊆ C _G (H _n ) ⊆ C _g(H) ⊆ N _g (H). If N ⊆ H _n , take N ₁ ⊲ N minimal normal in H _n and consider the conjugates N ^g₁ , g ∈ G.]

26. A finite group is nilpotent if and only if it contains a normal subgroup for each divisor of the order.

27. Prove that in a finite group G there always exists a nilpotent subgroup K such that G = 〈K ^g, g ∈ G〉.

28.

• i) Prove that a group is residually nilpotent if and only if ∩  _i ≥ o Γ _i  = {1}

• ii) prove that the infinite dihedral group (Ex. 2.10, 3) is residually nilpotent. On the other hand, this group admits nonnilpotent images.

5.3 p-Nilpotent Groups

Theorem 5.25 (Burnside’s basis theorem).

Let P be a finite p-group, and let Φ = Φ(P). Then:

• i) P/Φ is elementary abelian, and therefore a vector space over F _p ;

• ii) the minimal systems of generators of P all have the same cardinality d, where p ^d = |P/Φ|;

• iii) an element x of P, not belonging to Φ is part of a minimal system of generators of P.

Proof.

• i) The quotient P/Φ is abelian because P′ ⊆ Φ. Moreover, if M is maximal then P/M has order p and therefore (Mx)^p = M, i.e. x ^p ∈ M, for all M. Hence x ^p ∈ Φ, so all the elements of P/Φ have order p. (This could also be seen by observing that P/Φ = P/∩M _i ≤ P/M ₁ × P/M ₂ × P/M _k = Z ₁ × Z ₂ ×…× Z _k , where the M _i are all the maximal subgroups of P.)

• ii) By i), P/Φ is elementary abelian, so is generated by d elements, and not less. Hence P cannot be generated by less than d elements either, otherwise this would also be true for P/Φ.

• iii) If x ∉ Φ, then Φ x is a nonzero vector of P/Φ, so is part of a basis of this vector space: P/Φ = 〈Φ x ₁, Φ x ₂, …, Φ x _d 〉, where x ₁ = x. It follows P = 〈Φ, x ₁, x ₂,…, x _d 〉 = 〈x ₁, x ₂,…, x _d 〉. ◊

In general, P/Φ is distinct from the product of the P/M _i (think of the Klein group). Moreover, if the group is not a p-group the theorem is no longer true. The group C ₆ = 〈a〉 admits, besides {a}, of cardinality 1, the minimal generating system {a ², a ³}, of cardinality 2.

Lemma 5.5.

Let x ₁, x ₂, …, x _m be a generating system of a group G, and let a ₁, a ₂,…, a _m be elements of Φ = Φ(G). Then a ₁ x ₁, a ₂ x ₂, …, a _m x _m is also a generating system of G.

Proof. From G/Φ = 〈Φ x ₁, Φ x ₂,…,Φ x _m〉 and Φ x _i = Φ a _i x _i it follows

So G = 〈Φ, a ₁ x ₁, a ₂ x ₂,…,a _m x _m 〉 = 〈a ₁ x ₁, a ₂ x ₂,…,a _m x _m 〉. ◊

Theorem 5.26.

Let σ be an automorphism of a finite group G inducing the identity on G/Φ, where Φ = Φ (G) (i.e. x ¹ x ^σ ∈ Φ for all x ∈ G). Then the order of σ divides |Φ|^d, where d is the cardinality of a generating system of the group G.

Proof. Let G = 〈x ₁,x ₂, …,x _d 〉,. By the previous lemma, the components of the ordered d-tuples (a ₁ x ₁, a ₂ x ₂,…, a _d x _d), a _i ∈ Φ, also generate G. As the a _i vary in all possible ways in Φ we have a set Ω of |Φ|^d d-tuples. Now the group 〈σ〉 acts on this set according to (a ₁ x ₁, a ₂ x ₂,…, a _d x _d )^σ = ((a ₁ x ₁)^σ, (a ₂ x ₂)^σ, …, (a _d x _d )^σ); this is actually an action on Ω because by assumption x ^σ _i x ^− 1 _i  ∈ Φand therefore x ^σ _i  = a′ _i x _i , with $\begin{array}{cc}\hfill {a^{\prime}}_i\in \varPhi, \hfill & \hfill {\left({a}_i{x}_i\right)}^{\sigma }={a}_i^{\sigma }{x}_i^{\sigma }={a}_i^{\sigma }{a^{\prime}}_i{x}_i={a_i}^{\prime \prime }{x}_i\hfill \end{array}$, where a _i ^′ ′ ∈ Φ. If an element 〈σ〉 fixes a d-tuple, then it fixes all its components, and so all the elements of a generating system, and therefore is the identity. Thus the stabilizer of an element of Ω is the identity, i.e. 〈σ〉 is semiregular on Ω, so its orbits all have the same cardinality o(σ); hence |Ω| is a multiple of o(σ). ◊

Corollary 5.11.

If an automorphism σ of a finite p-group P induces the identity on P/Φ (P), then o(σ) is a power of p.

The latter corollary applies in particular when the p-group is a Sylow p-subgroup S of a finite group G. If g ∈ N _G (S), then g induces by conjugation an automorphism σ of S whose order divides o(g). σ is also an automorphism of S/Φ(S), and if it is trivial its order is a power of p. If g is a p′-element, then p ∤ o(σ). The only possibility is o(σ) = 1, and since g acts by conjugation this means that g centralizes S. Let us see a few examples of this situation.

We recall that an element g of finite order of a group is a product g = x ₁ x ₂ … x _m where the x _i are commuting p _i -elements, p _i ≠ p _j if i ≠ j,

Theorem 5.27.

Let S be a Sylow p-subgroup of a finite group G, and let S ∩ G′ ⊂ Φ(S). Then N _G (S) = S · C _G (S).

Proof. Let g ∈ N _G(S), and let g = xy where x is a p-element and y a p′- element. Let σ be the automorphism of S induced by conjugation by y. For all z ∈ S we have

Then by assumption z ⁻¹ z ^σ ∈ Φ(S), and therefore the automorphism σ induced by y on S is the identity. In other words, y centralizes S and so g = xy with x ∈ S and y ∈ C _G (S). ◊

Definition 5.7.

Let S be a Sylow p-subgroup of a group G. A subgroup K of G such that S ∩ K = {1} and G = SK is called a p-complement. If G has a normal p-complement then G is p-nilpotent.

A normal p-complement is unique. Indeed, it contains all the Sylow q-subgroups with q ≠ p, and therefore it consists of the elements of G of order coprime to p.

If a Sylow p-subgroup S is abelian, and S ∩ G′ ⊆ Φ(S), the previous theorem implies N _g (S) = C _G (S). As we shall see in Theorem 5.30, this condition implies the existence of a normal p-complement K. This subgroup K will be determined as the kernel of a homomorphism called the transfer, that we now define.

Let H be a subgroup of finite index n of a group G, and let x ₁, x ₂, …, x _n be representatives of the right cosets of H .If g ∈ G, then x _i g ∈ Hx _j for some j, and the mapping x _i → x _j thus obtained is a permutation σ = σ _g of the x _i:

where h _i ∈ H. In this equality, not only the x _σ ( _i ) but also the h _i are uniquely determined by g and x _i .

Theorem 5.28.

Let H be abelian. The mapping V : G → H given by g → ∏  ⁿ _i = 1 h _i , where the h _i are determined by g as in (5.12), is a homomorphism.

Proof. Let g, g′ ∈ G, and let x _i be as above and x _i g′ = h _i ′x _T (i). If x _i g = h _i x _σ (i), then

so that

where the second equality follows from the fact that a being a permutation, the set of the h′ _σ(i) coincides, as i varies, with the set of the h _i ′. ◊

Definition 5.8.

The homomorphism V : G → H of Theorem 5.28 is called the transfer ⁶ of G in H.

Remark 5.2.

The transfer does not depend on the choice of the representatives x _i of the right cosets of H in G (cf. ex. 30).

Now let x _i g = h _i x _σ ( _i ), and let (i ₁, i ₂,…, i _r−i , i _r ) be a cycle of σ. Then ${x}_{i_k}g={h}_{i_k}{x}_{i_{k+1}},k=1,2....r$ (indices mod r), and the contribution of these h _ik to the product Π h _i is ${h}_{i_1}{h}_{i_2}\dots {h}_{i_{r-1}}{h}_{i_r}$, that is,

If σ has t cycles, each of length r _i , let us choose for each cycle an element x _i such that such that

where ∑  ^t _i = 1 r _i  = n = [G : H]. Note that r _i is the least integer such that ${x}_i{g}^{r_i}{x}_r^{-1}\in H$.

Let us now see two applications of the transfer. The first one is Schur’s theorem, that we have already proved in a different way (Theorem 2.38); the second is a theorem of Burnside (Theorem 5.30).

Theorem 5.29 (Schur).

If the center of a group G is of finite index, then the derived group G′ is finite.

Proof. G′ is finitely generate (by a set of representatives of the cosets of the center). Now, setting Z = Z(G), we have G′/( G′∩ Z) ≃ G′Z/Z ≤ G/Z and therefore G′/ G′ ∩ Z has finite index in G′, and as such is finitely generated. It will be enough to show that it is torsion because, being abelian, it will be finite. Consider the transfer of G in Z .If g ∈ G′ ∩ Z, then ${x}_i{g}^{r_i}{x}_i^{-1}={g}^{r_i}$ and (5.13) becomes

where n = [G : Z]. The image of V being abelian, we have G′ ⊆ ker(V), and g ⁿ = 1 since g ∈ G′. ◊

Lemma 5.6 (Burnside).

⁷ Two elements of the center of a Sylow p-subgroup S of a group G that are conjugate in G are conjugate in the normalizer of S.

Proof. Let x, y∈ Z(S), and let y = x ^g. Then y belongs to the center of S and S ^g, so these two subgroups are Sylow in the centralizer C _G (y) of y and therefore S = (S ^g)^h, h ∈ C _G (y). Then gh ∈ N _G (S), and x ^gh = (x ^g)^h = y ^h = y, as required. ◊

Theorem 5.30 (Burnside).

Let G be a finite group, S a Sylow p-subgroup of G, and let N _G (S) = C _G (S) (i.e. S is contained in the center of its own normalizer). Then G has a normal p-complement.

Proof. S is abelian, since S ⊆ N _G (S) = C _G (S) and therefore is contained in its centralizer. The subgroup K we look for will be the kernel of the transfer V of G in S. Consider (5.13) with 1 ≠ g ∈ S. The elements ${g}^{r_i}$ and ${x}_i{g}^{r_i}{x}_i^{-1}$ are elements of S that are conjugate in G, and since S coincides with its center, by Lemma 5.6 they are conjugate in the normalizer of $S:y{g}^{r_i}{y}^{-1}={x}_i{g}^{r_i}{x}_i^{-1}$ with y ∈N _G (S). But being N _G (S) = C _G (S), y commutes with g ^ri, and therefore g ^ri = x _i g ^ri x ⁻¹. Then (5.13) becomes:

However g is a p-element, and n = [G : S] being coprime to p we have V(g) ≠ 1; thus ker(V) ∩ S = {1}. Let K = ker(V); then V(S) = SK/K≃S/S ∩K ≃ S, and V(S) = S. A fortiori, V(G) = S so G/K ≃ S. It follows G = SK with S ∩ K = {1}, as required. ◊

Let us now consider a few examples of application of Theorem 5.30.

Theorem 5.31.

• i) Let G be a finite group, and let p be the smallest prime divisor of the order of G. Assume that a Sylow p-subgroup S is cyclic. Then G has a normal p-complement.

• ii) Let S be a cyclic Sylow p-subgroup of a group G. Then either S∩G′ = S or S∩G′ = {1}.

Proof.

• i) Let us show that N = N _G (S) and C = C _G (S) coincide. We know that |N/C| ≤ |Aut(S)|. If |S| = p ⁿ, |Aut(S)| = φ(p ⁿ) = p ⁿ⁻¹(p − 1). S being abelian, S ⊆ C, so p ⁿ divides C. Hence |N/C| is coprime to p, and therefore divides p − 1. The prime p being the smallest divisor of |G| it follows |N/C| = 1 and N = C.

• ii) If S ⊆ G′, S ∩ G′ is a proper subgroup of S and therefore is contained in a maximal subgroup of S. Being a cyclic p-group S has a unique maximal subgroup, which therefore coincides with the Frattini subgroup. Hence S∩G′ ⊆ Φ(S). By Theorem 5.27, N _g (S) = S·C _g (S), and being S ⊆ C _g (S) because S is abelian, N _g (S) = C _g (S). By Burnside’s theorem, G has a normal p-complement: G = SK, K ⊴ G, S ∩ K = {1}. Since G/K ≃ S is cyclic, we have G′ ⊆ K, and S ∩ G′ = {1}.

S∩G′ is the focal subgroup of S in G. The quotient S/(S∩G′) is isomorphic to the largest abelian p-factor group of G (cf. ex. 40).

Theorem 5.32.

If G is a finite simple group, then either 12 divides |G|, or p ³ divides |G|, where p is the smallest prime divisor of |G|.

Proof. Let p be the smallest prime divisor of |G| and let |S| = p or p ². S is abelian. If it is cyclic, by the previous theorem G is not simple; if it is elementary, then |N/C|< |Aut(S)| = (p ² − 1)(p ² − p) = p(p − 1)²(p + 1). If p> 2, no prime greater than p divides the latter integer, so no prime divides |N/C|; it follows |N/C| = 1 and N = C .If p = 2, S is a Klein group, and Aut(S) is isomorphic to S ³. If N/C ≠ {1}, since V ⊆ C we have that 4 divides |C| and therefore |N/C| = 3. Hence |G| is divisible by 4 and by 3, and so by 12. ◊

Example 5.5.

Let us prove that A ⁵ is the only simple group of order p ² qr, p < q < r. By the previous theorem, the order of the group must be divisible by 12, hence p = 2, q = 3 and the order is 12r. We have already seen that the unique prime number r compatible with simplicity is 5 (Chapter 3, ex. 43). Hence the order of the group is 60, and G ≃ A⁵ (Ex. 3.5, 3).

5.4 Exercises

29.

• i) If a group is p-nilpotent for all primes p, then it is nilpotent;

• ii) if a group is p-nilpotent, then so also are subgroups and factor groups.

30. Prove that the transfer does not depend on the choice of a transversal.

31. G be p-nilpotent, and let N be a minimal normal subgroup of G with p||N|. Prove that N is a p-group contained in the center of G.

32. Let G be p-nilpotent, and let G = SK. Prove that Z(S)K ⊴ G.

33. If S, S ₁ ∈ S _y l _p (G), then Z(S) ⊴ S ₁ implies Z(S) = Z(S ₁).

34. G is said to be p-normal if given two Sylow p-subgroups S and S ₁, then Z(S) ⊆ S ₁ implies Z(S) = Z(S ₁). Prove that:

• i) if the Sylow p-subgroups are abelian, then G is p-normal;

• ii) if any two Sylow p-subgroups have trivial intersection, then G is p-normal;

• iii) if G is p-nilpotent, then G is p-normal;

• iv) S ⁴ is not 2-normal.

  35. Let G be a finite group such that the normalizer of every abelian subgroup coincides with its centralizer. Prove that G is abelian. [Hint: prove that the Sylows are abelian by considering an abelian maximal subgroup of a Sylow p-subgroup and then prove that G is p-nilpotent by applying Burnside’s theorem.]

36. Let S be a Sylow p-subgroup of order p of a group G, and let x ≠ 1 be an element of S. Prove that:

• i) under the action of G on its Sylow p-subgroups by conjugation, the permutation induced by x consists of a fixed point and of cycles of length p;

• ii) an element of G that normalizes but does not centralize S cannot fix two elements belonging to the same orbit of S, and therefore the number of Sylow p-subgroups it fixes is at most equal to the number of orbits of S.

37. Use i) of the previous exercise to prove that groups of order 264, 420 or 760 cannot be simple.

38. Using ex. 36 and (3.9) prove that a group of order 1008 or 2016 cannot be simple.

39. If p ², p a prime, divides |G|, then G has an automorphism of order p.

40. Let S ∈ S _y l _p (G). Prove that:

• i) if H ⊴ G is such that G/H is an abelian p-group, then S ∩ G′ ⊆ H and G/H is isomorphic to a homomorphic image of S/ (S ∩ H);

• ii) there exists a subgroup H ⊴ G such that G/H is isomorphic to S/(S ∩ H) [Hint: consider the inverse image H of O ^p(G/G′) (cf. Chapter 3, ex. 32); also S ∩ G′ = S ∩ H.]

5.5 Fusion

If two elements of a finite group are conjugate, then two suitable powers of them are p-elements, for some p, and are also conjugate. Since any two Sylow p-subgroups are conjugate, a conjugation of these two powers reduces to that of two elements belonging to the same Sylow p-subgroup. In this section, we shall be considering the converse problem: when are two elements of a Sylow p-subgroup S that are not conjugate in S conjugate in G?

Definition 5.9.

Two elements or subgroups of a Sylow p-subgroup S of a group G are said to be fused in G if they are conjugate in G but not in S. (The two conjugacy classes of S to which the two elements or subgroups belong fuse into one conjugacy class of G.)

We have already seen two results about fusion: one is Lemma 5.6 of Burn- side, concerning the conjugation of two elements of the center of a Sylow p-subgroup. The other one, again by Burnside, is Lemma 3.8, concerning conjugation in the group of two normal subgroups of a Sylow p-subgroups. (This result extends immediately to the case of two normal sets, that is, sets that coincide with the set of conjugates of their elements.) If the normality assumption is dropped, two subsets of S, although not necessarily conjugate in the normalizer of S any more, are conjugate by an element that is a product of elements belonging to normalizers of subgroups of S. This is the content of a theorem of Alperin that we shall prove in a moment.

Lemma 5.7.

Let S be a Sylow p-subgroup of a group G, and let T be a subgroup of S. Then there exists a subgroup U of S which is conjugate to T in G and such that N s(U) is Sylow in N g(U).

Proof. T is contained in a Sylow p-subgroup P of its normalizer N _G (T), and there exists g ∈ G such that P ^g ⊆ S. The subgroup we seek is U = T ^g. Indeed, P ^g is Sylow in N _G (T)^g =N _G (T ^g) = N _G (U), is contained in S, and therefore is contained in N _S (U), but since it contains a Sylow p-subgroup of N _g (U) it coincides with it. ◊

Definition 5.10.

Let S be a Sylow p-subgroup of a group G, ℱ a family of subgroups of S. Let A and B be two nonempty subsets of S and let g be an element of S. Then A is said to be ℱ-conjugate to B via g if there exist subgroups T ₁, T ₂,…, T _n of the family ℱ and elements g ₁, g ₂, …, g _n of G such that:

• i) g _i ∈ N g(T _i ), i = 1, 2,…,n;

• ii) 〈A〉 ⊆ T₁, ${\langle A\rangle}^{g_1,{g}_2\dots {g}_i}\subseteq {T}_{i+1}$, i = 1, 2,…,n−1;

• iii) A ^g = B, where g = g ₁ g ₂… g _n .

The family ℱ is called a conjugation family (for S in G) if, given any two subsets A and B of S that are conjugate in G by an element g, then A is F- conjugate to B by g.

Lemma 5.7 suggests how to determine a conjugation family.

Theorem 5.33 (Alperin).

Let S be a Sylow p-subgroup of a group G, and let ℱ the family of subgroups T of S such that N _S (T) is Sylow in N _g (T). Then ℱ is a conjugation family for S in G.

Proof. Let A, B ⊆ S, A ^g = B; we prove that A is ℱ-conjugate to B via g. By induction on the index of 〈A〉 in S. Setting T = 〈A〉 and V = 〈B〉, we have T ^g = V. Let [S : T] = 1; then S = T, S ^g = T ^g = V ⊆ S, and therefore S ^g = S, i.e. g ∈ N _g (S). We obtain i), ii) and iii) of Definition 5.10 by putting n =1, T ₁ = S and g ₁ = g, and by observing that S certainly belongs to ℱ. Now let [S : T] > 1, i.e. T < S. Then T < N _S (T) and V < N _S (V). Let U ∈ ℱ be conjugate to T ^g (Lemma 5.7); then there exists h ₁ ∈ G such that, from which ${N}_S{(T)}^{g{h}_1}\subseteq {N}_G{(T)}^{g{h}_1}={N}_G\left({T}^{g{h}_1}\right)={N}_G(U)$. Since N _G (U) is Sylow in N _G (U), there exists h ₂ ∈ N _G (U) such that ${N}_S{(T)}^{g{h}_1g{h}_2}\subseteq {N}_S(U).$ Setting h = h ₁ h ₂, we have T ^gh = U, N _S (T)^gh ⊆ N _S (u). Similarly, there exists k ∈ G such that V ^k = U, N _S (T)^k ⊆ N _S (U). The index of N _S (T) in S being less than that of T in S, by induction N _S (T) is ℱ-conjugate to N _S (T)^gh via gh, and N _S (V)^k is ℱ-conjugate to N _g (V) via k ⁻¹. Then (ex. 46) A is ℱ-conjugate to B ^h via gh and B ^k is ℱ-conjugate to B via k ⁻¹. Moreover, since h ⁻¹ k ∈ N _G (U) and U ∈ℱ, B ^h is ℱ-conjugate to B ^k via h ⁻¹ k. From the sequence of ℱ-conjugations:

and by ex. 47, A is ℱ-conjugate to B via (gh)(h ⁻¹ k)k ⁻¹ = g. ◊

This theorem may be expressed by saying that conjugation has a local character. Normalizers of nonidentity p-subgroups are called p-local subgroups. Thus, Alperin’s theorem may be expressed by saying that two elements or subsets of a Sylow p-subgroup S of a group G are conjugate in G if and only if they are locally conjugate in G.

5.6 Exercises

41. Two element of the centralizer of a p-Sylow S that are conjugate in G are conjugate in the normalizer of S.

42. Let S be Sylow in G, and let H ≤ G be such that H ^g ⊆ S implies H ^g = H (H is said to be weakly closed in S w.r.t. G). Prove that two elements of the centralizer of H that are conjugate in G are conjugate in the normalizer of H. (Note that since H = S is certainly weakly closed in S, this exercise generalizes the case of two elements of the center of a Sylow (Lemma 5.6) already generalized in the previous exercise.

43. Let the p-Sylows of a group G have pairwise trivial intersection. Prove that two elements x and y of a p-Sylow S that are conjugate in G are conjugate in the normalizer of S. In fact, any element that conjugates them belongs to Ng(S).

44. Let G be p-nilpotent. Then:

• i) two p-Sylows are conjugate via an element of the centralizer of their intersection;

• ii) if S is a p-Sylow, two elements of S that are conjugate in G are conjugate in S ⁸;

• iii) if P is a p-subgroup of G, then N _G (P)/C _G (P) is a p-group.

45. If (n, φ(n)) = 1, then all group of order n are cyclic. Conversely, if all groups of order n are cyclic, then (n, φ(n)) = 1.

46. Let ℱ be a family of subgroups of a p-Sylow S of a group G, and let A and B be two nonempty subsets of S with A ℱ-conjugate to B via g. If C is a nonempty subset of A, prove that C is ℱ-conjugate to C ^g via g.

47. With S and ℱ as in the previous exercise, let A ₁, A ₂, …, A _m+1 be subsets of S, and let h ₁, h ₂, …, h _m be elements of G such that A _i is ℱ-conjugate to A _i+1 via h _i , i = 1, 2,…,m. Prove that A _i is ℱ-conjugate to A _m+1 via the product h ₁ h ₂…h _m.

48. Consider the simple group G of order 168. In the dihedral Sylow 2-subgroup S let x and y be two involutions that generate S. These two involutions are not conjugate in S. They are conjugate in G (all involutions are conjugate in G), but not in Ng(S) (because the latter equals S). Hence there is no subgroup H of S containing x and y such that the two elements are conjugate in the normalizer of H. Prove that if z be the central involution of S, then x and z are conjugate in the normalizer of the Klein group they generate, and the same happens for z and y. The product of the two conjugating elements conjugates x and y.

5.7 Fixed-Point-Free Automorphisms and Frobenius Groups

Definition 5.11.

An automorphism σ of a group is fixed-point-free (f.p.f.) if it fixes only the identity of the group.

Theorem 5.34.

Let a be a f.p.f. automorphism of a finite group G. Then:

• i) the mapping G → G given by x → x ⁻¹ x ^σ (or x → x ^σ x ⁻¹) is one-to-one (but not an automorphism, in general);

• ii) if N is a normal σ-invariant subgroup G, σ acts f.p.f. on G/N;

• iii) for each prime p dividing |G| there exists one and only one Sylow p-subgroup of G fixed by σ.

Proof.

• i) If x ⁻¹ x ^σ = y ⁻¹ y ^σ, then (yx ⁻¹)^σ = yx ⁻¹ and therefore yx ⁻¹ = 1 and y = x. The mapping is injective, and since the group is finite also surjective.

• ii) If (Nx)^σ= Nx x ⁻¹ x ^σ ∈ N. As to i), there exists h ∈ N such that x ⁻¹ x ^σ = h ⁻¹ h ^σ; as above x = h, so that Nx=N.

• iii) If S is a p-Sylow, S ^σ= S ^x for some x ∈ G, and by i) we have x = y ^-1 y ^σ. Hence S^σ = S ^x = (y ⁻¹)^σ ySy ⁻¹ y ^σ, from which y ^σ S ^σ (y ⁻¹) ^σ = ySy ⁻¹, or (ySy ⁻¹) ^σ = ySy ⁻¹, and the latter is a p-Sylow fixed by σ. If σ fixes S and S ^x, then ${S}^x={\left({S}^x\right)}^{\sigma }={\left({S}^{\sigma}\right)}^{x^{\sigma }}={S}^{x^{\sigma }}$, from which x ^σ x ⁻¹ ∈ N _G (S). Since N _g (S) is σ-invariant if S is, there exists y ∈ N _G (S) such that x ^σ x ⁻¹ = y ^σ y ⁻¹. But then x = y and x ∈ N _G (S) so that S ^x = S. ◊

In Theorem 5.31, i), we have seen that if a Sylow p-subgroup of a group G is cyclic, where p is the smallest prime dividing the order of G, then G has a normal p-complement. The next theorem shows that the minimality assumption on the prime p may be replaced with G admits a f.p.f. automorphism. First a lemma.

Lemma 5.8.

Let σ be a f.p.f. automorphism of a group G, and let H be a normal cyclic and a-invariant subgroup. Then H ⊆ Z(G).

Proof. H being cyclic, Aut(H) is abelian, hence the restriction of σ to H (still denoted by a) commutes with all the elements of Aut(H), and in particular with those induced by conjugation by the elements of G: h ^σγg = h ^γgσ, for all h ∈ H .It follows g ⁻¹ h ^σ g = (g ^σ)⁻¹ h ^σ g ^σ, that is g ^σ g ⁻¹ commutes with all the h ^σ, and so with all h ∈ H: g ^σ g ⁻¹ ∈ C _G (H). C _g (H) being σ-invariant, there exists x ∈ C _g (H) such that g ^σ g ⁻¹ = x ^σ x ⁻¹, so g = x. Then every g ∈ G centralizes H, and the result follows. ◊

Theorem 5.35.

Let G be a group admitting a f.p.f. automorphism a and let a Sylow p-subgroup be G cyclic. Then G ha a normal p-complement.

Proof. By Theorem 5.34, iii), there exists a σ-invariant Sylow p-subgroup whose normalizer N is also σ-invariant. By the lemma above S ⊆ Z(N), and by Burnside’s theorem (Theorem 5.30), G has a normal p-complement. ◊

The notion of a f.p.f. automorphism presents itself naturally when dealing with a special class of groups, the Frobenius groups.

Definition 5.12.

A finite transitive permutation group on a set Ω in which only the identity fixes more than one point, but the subgroup fixing a point is non trivial, is called a Frobenius group.

Example 5.6.

The dihedral group D _n , n odd, viewed as a permutation group of the vertices of a regular n-gon, is a Frobenius group. It is transitive, and a vertex is fixed only by the symmetry w.r.t. the axis passing through that vertex. The remaining symmetries are rotations, so fix no vertex. If n is even, the symmetry w.r.t. the axis passing through two opposite vertices fixes these two vertices, so this group is not a Frobenius group. See also Ex. 5.7.

Theorem 5.36.

Let G be a Frobenius group. Then:

• i) the elements of G fixing no point are regular permutations, and they are n − 1 in number;

• ii) a permutation fixing a point is regular on the the remaining ones, so the subgroup H = G _α fixing a point is semiregular on n − 1 points. Hence its order divides n − 1;

• iii) N _G (H) = H.

Proof.

• i) The cycles of a permutation a fixing no point have length at least 2, and if there are cycles of length h and k, with h > k, then the k-th power of this permutation fixes at least k ≥ 2 elements (those of the k-cycle), and this power is not the identity because the k-th power of the h-cycle is not the identity. Now, and since G _α ∩ G _β = {1} the elements fixing no point are those of the set $G\backslash \left(\underset{\alpha \in \varOmega }{\cup }{G}_{\alpha}\right)=\left\{1\right\}.$ G being transitive, $\left[G:{G}_{\alpha}\right]=n,\mathrm{so}\;\left|\underset{\alpha \in \varOmega }{\cup {G}_{\alpha }}\right|=\left(\left|{G}_{\alpha}\right|-1\right)n,$ and the result follows:

• ii) follows from i);

• iii) the elements of H ^g fix α ^g, so if H ^g = H then α ^g = α that is g ∈ H. ◊

Theorem 5.37.

Let G be a finite group containing a subgroup H disjoint from its conjugate and self normalizing:

• i) H ∩ H ^x = {1}, x ∉ H;

• ii) N _G ( H) = H.

Then G is faithfully represented as a Frobenius group by right multiplication on the cosets of H.

Proof. In the action of G on the cosets of H the stabilizer of a coset Hx is the subgroup H ^x and therefore is not the identity. If g ∈ G fixes two cosets Hx and Hy, then g ∈ H ^x ∩ H ^y, and by i) either g = 1 or H ^x = H ^y, i.e. xy ⁻¹ ∈ N _G (H) = H and Hx = Hy. ♦

The mapping φ : Hg → α ^g establishes an equivalence between the action of the group G on the cosets of H, and that of G on the set Ω of Definition 5.12 with H = G _α . (φ is well defined because H fixes α.) In other words for a finite group G the following conditions:

• i) G acts transitively on a set and in this action only the identity fixes more than one point, but the subgroup fixing a point is nontrivial;

• ii) G is a group containing a subgroup H disjoint from its conjugate and self normalizing,

are equivalent. Hence we may define a Frobenius group as a group satisfying either condition.

The elements fixing no point form, together with the identity, a subgroup K. This is the content of a theorem of Frobenius, that we shall prove in the next chapter (Theorem 6.21). The subgroup K is normal (two conjugate elements fix the same number of points (zero, in this case).

Definition 5.13.

The subgroup H of a Frobenius group is a Frobenius complement. The subgroup K is the Frobenius kernel.

Theorem 5.38.

The Frobenius kernel of a Frobenius group G is regular.

Proof. K being semiregular (Theorem 5.36, i)), it suffices to show that it is transitive. By Theorem 5.36, i), |K| = n. The index of the stabilizer of a point in K being the identity, the orbit of a point contains |K| = n points, so K is transitive. ◊

Theorem 5.39.

Let G be a Frobenius group with complement H and kernel K. Then:

• i) G = HK, with H ∩ K = {1} and K ⊴ G, so that G is the semidirect product of K by H;

• ii) |H| divides |K| − 1.

Proof.

• i) Follows from the above quoted Frobenius theorem and from the fact that the transitivity of K implies G = HK.

• ii) This is point ii) of Theorem 5.36, recalling that |K| = n. ◊

Theorem 5.40.

In a Frobenius group,

• i) the conjugacy class in G of a non identity element h ∈ H coincides with the conjugacy class of h in H;

• ii) the centralizer of an element 1 ≠ h ∈ H is contained in H, so C _G (h) = C _h (h);

• iii) an element not in K is conjugate to an element of H. Hence a set of representatives of the conjugacy classes of H is also a set of representatives of the conjugacy classes of thee elements not in K;

• iv) the centralizer of an element 1 ≠ k ∈ K of K is contained in K, so C _G (k) = C _k (k).

Proof.

• i) Let h = g ⁻¹ h ₁ g, for h, h ₁ ∈ H and g ∈ G. Then h ∈ H ∩ H ^g, so h fixes both α and α ^g, forcing α = α ^g and g ∈ H;

• ii) the equality of the centralizers follows by taking h′ = h in i),

• iii) an element not in K belongs to the stabilizer of a point which is conjugate to H,

• iv) if g ∈ C _G (k) then, by ii), g fixes no element of a conjugate of H. But the conjugates of H are the stabilizers of points, so g ∈ G _α for some α, and again by ii) it cannot commute with an element of K. ◊

By iii) of this theorem the nonidentity elements of H induce by conjugation f.p.f. automorphisms of K. The next theorem shows that this fact characterizes Frobenius groups among semidirect products.

Theorem 5.41.

Let G = HK with K ⊴ G be a semidirect product, and suppose that thee action of H on K is a f.p.f. action. Then G is a Frobenius group with kernel K and complement H.

Proof. Let us show that H is disjoint from its conjugate and is self-normalizing. The result will follow from Theorem 5.37. If g ∈ G, then g = hk, so H ^g = H ^k. Let H ∩ H ^k ≠ {1}; then h ₁ = k ⁻¹ h ₂ k, and multiplying by h ^− 1₂ we obtain h ₁ h ^− 1₂  = k ^− 1 h ₂ kh ^− 1₂  = k ^− 1(h ₂ kh ^− 1₂ ) ∈ K, by the normality of K. Hence h ₁ h ^− 1₂  ∈ H ∩ K = {1} and h ₁ = h ₂. This means that k commutes with h ₁, and the fact that the action is f.p.f. implies k =1 and H ^k = H. Hence, if g ∉ H then H ∩ H ^g = {1}, and moreover H ^g ≠ H if g ∉ H, i.e. no element outside H normalizes H. ◊

Lemma 5.9.

If G is a finite group admitting a f.p.f. automorphism σ of order 2. Then G is abelian and of odd order.

Proof. If x ∈ G, then x = y ⁻¹ y ^σ for some y ∈ G. It follows

Then σ is the automorphism x → x ^–1, and a group admitting such an automorphism is abelian. Moreover, σ being f.p.f., if x ≠ 1 we have x ≠ x ⁻¹, therefore there are no elements of order 2. Hence G is of odd order. ◊

Theorem 5.42.

Let G be a Frobenius group with complement H of even order. Then the kernel K is abelian.

Proof. An element of order 2 of H induces a f.p.f. automorphism of K of order 2. Apply the previous lemma. ◊

Examples 5.7.

• 1. (Cf. Ex. 2.10, 2) Let K be a finite field. The group $\mathcal{A}$ of affine transformations over K is defined, as in the case of the field of real numbers, as the semidirect product of the subgroup $\mathcal{A}$ ₀ of the homotheties and the normal subgroup of the translations $\mathcal{T}$. With ψ _a a homothety φ _b and a translation, let ψ ^− 1 _a φ _b ψ _a  = φ _b . Then x + a ⁻¹ b = x + b, so either b = 0, and φ _b is the identity, or a =1, and ψ _a is the identity. Hence $\mathcal{A}$ ₀ acts fixed-point-freely on the subgroup of translations, so $\mathcal{A}$ is a Frobenius group with kernel $\mathcal{T}$ and complement $\mathcal{A}$ ₀. The subgroup $\mathcal{A}$ ₀ is the stabilizer of 0; the other complements are the stabilizers of the nonzero elements of K . If s ≠ 0, then s is fixed by the transformations ϕ _a,s−as as a varies in K, and these form a subgroup A _s which is conjugate to $\mathcal{A}$ ₀ via the translation φ _s . If an element ϕ ∈ A _s fixes t, then ϕ(t) = at + s − as = t, that is (a − 1)s = (a − 1)t. If a ≠ 1, then s = t, and if a =1, then ϕ(x) = x, for all x, and therefore ϕ =1. If an affinity ϕ _a,b is not a translation, then a ≠ 1 and ϕ _a,b fixes $\frac{b}{1-a}.$ Hence the translations are the only affinities fixing no point. The group is transitive because already $\mathcal{T}$ is transitive. $\mathcal{A}$ ₀ being isomorphic to the multiplicative group of K, if |K| = q we have |$\mathcal{A}$| = q − 1, and $\mathcal{T}$ being isomorphic to the additive group of K we have |$\mathcal{T}$| = q and |$\mathcal{A}$| = (q − 1)q.

• 2. A ⁴ is a Frobenius group with kernel a Klein group and complement a cyclic group. It can be seen as the group of affine transformations over the 4-element field K = {0, 1, x, x + 1}, with the product modulo the polynomial x ² + x + 1.

• 3. The cyclic group C ₅ = 〈a〉 admits the automorphism σ = (a, a ², a ⁴, a ³) of order 4, and the semidirect product C ₅〈σ〉 is isomorphic to the group of affine transformations over the field Z ₅. Indeed, the additive group of this field is generated by 1, and by identifying 1 with the translation τ ₁ : x → x +1, and conjugating it by the homothety ψ ₂, we have ψ ₂ Τ ₁ ψ ^− 1₂  : x → x + 2. In this way we have the automorphism of Z ₅(+) given by the cyclic permutation (1, 2,4, 3), and by letting correspond the latter to σ and i (or τ _i ) to a ⁱ we have the required isomorphism.

• 4. Let G be the semidirect product of C ₇ = 〈b〉 and C ₃ = 〈a〉, with the action of C ₃ on C ₇ given by a ⁻¹ ba = b ². The permutation induced by a on C ₇ is (1)(b, b ², b ⁴)(b ³, b ⁶, b ⁵). Only 1 is fixed by a, and therefore also by a ⁻¹; hence G is Frobenius. It has order 21, and is a subgroup of the simple group of order 168 (the normalizer of a 7-Sylow). Another group containing it is the group of affinities over the field Z ₇, of order 42, and is the semidirect product of C ₆ and Z ₇(+). The group G is the semidirect product of the C ₃ of C ₆ with Z ₇(+).

5.8 Exercises

49. If σ is f.p.f. and x ^σ is conjugate to x, then x = 1, that is, σ permutes the conjugacy classes of the group and fixes only the class of the identity. [Hint: Theorem 5.34, i).]

50. If σ is f.p.f o(σ) = n, then $x{x}^{\sigma }{x}^{\sigma^2}{x}^{\sigma^3}\cdots {x}^{\sigma^{n-1}}=1.$

51. If σ is f.p.f and o(σ) = 3, then x and x ^σ commute. (It can be proved that the group is nilpotent.)

52. If σ is f.p.f and I is the group of inner automorphisms of G, then the coset σI of I in Aut(G) consists of f.p.f. automorphisms, and these are all conjugate.

53. A p-subgroup P σ-invariant of G is contained in the unique p-Sylow S σ-invariant of G. [Hint: consider a subgroup H maximal w.r.t. the property of containing P and being σ-invariant, and prove that H = S.]

54. Two commuting f.p.f. automorphisms fix the same Sylow p-subgroup (cf. Theorem 5.34, iii)). In particular, if H is an abelian group of f.p.f. automorphisms of a group G, the elements of H all fix the same Sylow p-subgroup of G.

55. Let α ∈ Aut(G) (not necessarily f.p.f.) fixing only one p-Sylow S for all primes p. Then C _G (α) = {x ∈ G | x ^α = x} is nilpotent. [Hint: prove that C _G (α) ⊆ N _G (S).]

56. If G is infinite, Lemma 5.9 is no longer true. [Hint: consider D _∞ (Ex. 4.6, 3).]

57. Write down the 12 affinities of Ex. 5.7, 2.

58. Let G be a Frobenius group of kernel K and complement H. Prove that if N ⊲ G, and N ∩ H = {1}, then N ⊆ K.

5.9 Solvable Groups

A central series of a group G is an invariant series in which the quotient H/K is contained in the center of G/K, and therefore is abelian. Hence we may generalize the notion of a central series to that of an invariant series with abelian quotients. The next generalization, that of a normal series with abelian quotients is fictitious because it is not obtained a larger class of groups. This is what we shall see in a moment; it is related to the existence of the derived series, that we now define.

Definition 5.14.

The series of subgroups:

is the derived series of G. The subgroup G ⁽ⁱ⁺¹⁾ is the derived group of G ⁽ⁱ⁾, so the quotients G ⁽ⁱ⁾/G ⁽ⁱ⁺¹⁾ are abelian.

The subgroups G(i) are characteristic in G: if α is an automorphism of G = G ⁽⁰⁾ then (G ⁽⁰⁾)^α = G(0); by induction, (G ⁽ⁱ⁾)^α = G ⁽ⁱ⁾, and therefore (G ⁽ⁱ⁺¹⁾)^α = [G ⁽ⁱ⁾, G ⁽ⁱ⁾]^α = [(G ⁽ⁱ⁾)^α, (G ⁽ⁱ⁾)^α] = [G ⁽ⁱ⁾, G ⁽ⁱ⁾] = G ⁽ⁱ⁺¹⁾.

Theorem 5.43.

Let G = H ₀ ⊇ H ₁ ⊇ H ₂ ⊇ … ⊇ H ^(m−1) ⊇ H ^(m) = {1}, be a normal series with abelian quotients. Then G ⁽ⁱ⁾⊆ H ⁽ⁱ⁾, i =1, 2, ..., m. In other words, if a normal series with abelian quotients reaches the identity in m steps, the derived series reaches the identity in at most m steps.

Proof. By induction on m. We have G = G ⁽⁰⁾ ⊆ H ⁽⁰⁾ = G; assume G ⁽ⁱ⁾ ⊆ H ⁽ⁱ⁾. The quotient H _i /H _i+1 being abelian, we have H _i ′ ⊆ H _i + 1, and in particular (G ⁽ⁱ⁾)′ = G ^(i + 1) ⊆ H _i + 1 ⋅ ⋄

Theorem 5.44.

In a group consider:

• i) an invariant series with abelian quotients;

• ii) a normal series with abelian quotients;

• iii) the derived series.

Then if one of these series reaches the identity, then so do the other two.

Definition 5.15.

A group G is solvable if, for some n, G ⁽ⁿ⁾ = {1} in (5.14). The smallest such n is the derived length of G.

Examples 5.8.

1. An abelian group is solvable, with derived length 1. A nilpotent group is solvable: G′ = Γ ₂, and by induction

Polycyclic groups are solvable.

2. S ³ is solvable, with derived series S ³ ⊃ C ₃ ⊃ {1}, and so is S ⁴, with derived series S ⁴ ⊃ A ⁴ ⊃ V ⊃ {1}. The simplicity of A ⁿ, n ≥ 5, implies that S ⁿ is not solvable: the terms of the derived series coincide with A ⁿ from G′ = A ⁿ on.

3. A solvable simple group has prime order.

Theorem 5.45.

• i) Subgroups and quotient groups of a solvable group G are solvable, with derived length at most that of G;

• ii) if N ⊴ G and G/N are solvable, so is G.

Proof.

• i) If H ≤ G then H ⁽ⁱ⁾ ⊆ G ⁽ⁱ⁾, so if G ⁽ⁿ⁾ = {1} then H ⁽ⁿ⁾ = {1}. If N ⊴ G we have (G/N)⁽ⁱ⁾ = G ⁽ⁱ⁾ N/N, as follows easily by induction.

• ii) If (G/N)^(m)= N, then G ^(m) N/N = N and therefore G ^(m) ⊆ N. If N ^(r) = {1}, then G ^(m+r) = (G ^(m))^(r) ⊆ N ^(r) = {1}. ⋄

Remark 5.3.

As we know, ii) of the above theorem is not true if “solvable” is replaced by “nilpotent”.

Let now N be a minimal normal subgroup of a solvable group G. N being solvable, the derived group N′ is a proper subgroup of N, and as a characteristic subgroup is normal in G. By minimality, N′ = {1}, so N is abelian. If N is finite, let p be a prime dividing its order. The elements of order p of N form a subgroup, which is characteristic and nontrivial and therefore coincides with N. Hence N is an elementary abelian p-group. We have proved:

Theorem 5.46.

Let N be a minimal normal subgroup of a solvable group G. Then:

• i) N is abelian;

• ii) if N is finite, N is an elementary abelian p-group.

Hence, in a finite solvable group there always exists a non trivial normal p-subgroup H, for some prime p; in particular, the intersection of the Sylow p-subgroups for that prime p is nontrivial because it contains H. Moreover, H ⊆ F(G), so F(G) is nontrivial. Another nontrivial subgroup contained in F(G) is the second to the last term of a composition series.

Remark 5.4.

In a nilpotent group, finite or infinite, a minimal normal subgroup is a Z p. In in a finite solvable group a minimal normal subgroup is a direct sum of copies of Z p.

Corollary 5.12.

The chief factors of a finite solvable group are elementary abelian.

Corollary 5.13.

In a finite solvable group G a minimal normal subgroup is contained in the center of F(G). Moreover, a normal subgroup of G meets F(G) non trivially.

Proof. By Theorem 5.15, F(G) centralizes the minimal normal subgroups of G. If N is such a subgroup, N is a p-group, and therefore nilpotent. It follows N ⊆ F(G) and therefore N ⊆ Z(F(G)). If H ⊴ G, then H contains a minimal normal subgroup N; but N ⊆ F(G), so H ∩ F(G) ≠ {1}. ⋄

Remark 5.5.

It follows from the above that in finite solvable groups the Fitting subgroup plays the role the center plays in nilpotent groups.

If a solvable group has a composition series, the chief factors are solvable and simple, so are of prime order (in particular the group is finite). Conversely, a composition series with factor groups of prime order is a normal series with abelian factor groups reaching the identity; a group with such a series is solvable. Taking into account Corollary 5.12, we have the following characterization of finite solvable groups.

Theorem 5.47.

Let G be a finite group. Then the following are equivalent:

• i) G is solvable;

• ii) the chief factors of G are elementary abelian;

• iii) the composition factors are of prime order.

From iii) it follows that a finite solvable group always contains a normal subgroup of prime index (the second term of a composition series).

Corollary 5.14.

A solvable group admits a composition series if and only if is finite.

The fact that in a solvable group the Fitting subgroup is nontrivial may be used to construct an invariant series of a solvable group whose length may be considered as a measure of how far the group is from being nilpotent. Define inductively:

Hence F _i+1 is the inverse image of the Fitting subgroup F(G/F _i ) in the canonical homomorphism G → G/F _i . We have a series {1} = F ₀ ⊂ F ₁ ⊂ F ₂ ⊂..., in which if G is solvable the inclusions are strict (Corollary 5.13), and therefore it stops only when it reaches G.

Definition 5.16.

Let G be a finite solvable group. The series

is the Fitting series of G, and the integer n the Fitting length or Fitting height of G.

Examples 5.9.

• 1. F ₀ = G if and only if G = {1}; F ₁ = G if and only if G is nilpotent. Moreover, if F _n = G for a certain n, then G is solvable. Indeed, F ₁ is solvable because is nilpotent; assuming by induction that F _i is solvable, we have that F _i+1/F _i is nilpotent, hence solvable, from which it follows F _i+1 solvable (Theorem 5.45, ii)).

• 2. The Fitting series of S ⁴ is {1} ⊂ V ⊂ A ⁴ ⊂ S ⁴.

If K ⊆ H are two normal subgroups of a group G, the action of G on H/K induces a homomorphism G → Aut(H/K) with kernel C _G (H/K) (cf. Definition 5.5). Denoting the image by Aut _G (H/K) we have:

In Theorem 5.16 we saw that the Fitting subgroup is the intersection of the centralizers in G of the quotients of a chief series {H _i }, i = 1, 2,..., l, of G. It follows:

If G is solvable, the quotients H _i /H _i+1 are elementary abelian, hence they are vector spaces over a field with a prime number of elements (the primes may vary). Then Aut _G (H _i /H _i+1) is a linear group. Moreover, since there are no normal subgroups of G in between H _i and H _i+1, these spaces have no G-invariant subspaces: G is irreducible. Therefore, if G is a solvable finite group, the structure of G/F(G) is determined by that of the irreducible solvable subgroups of the finite dimensional linear groups over finite prime fields.

Let us now consider a few properties of the Fitting subgroup of a finite solvable group.

Theorem 5.48.

Let G be a finite solvable group, F = F(G) its Fitting subgroup. Then:

• i) C _G (F) = Z(F).

  Let F be cyclic; then:

• ii) G′ is cyclic;

• iii) if p ∤ |F|, a Sylow p-subgroup is abelian;

• iv) if F is a p-group, then p is the largest prime divisor of |G|.

Proof.

• i) Let H = F C _G (F). Then F(H) = F, C _H (F) = C _G (F), and therefore if H < G, by induction C _H (F) = Z(F(H)) and the claim follows. Then let G = F C _G (F); it follows:.

  

  and

  

  where the first equality follows by induction. Then C _G (F) ⊆ F and the claim follows.

• ii) F being abelian, F ⊆ C _G (F) = Z(F) ⊆ F, so C _G (F) = F. Since G/F = N _G (F)/C _G (F), which is isomorphic to a subgroup of Aut(F), which is abelian because F is cyclic, we have G′ ⊆ F, and so G′ is cyclic.

• iii) G/F being abelian, SF/F is also abelian, where S ∈ S _y l _p (G), and SF/F ≃ S/S ∩ F ≃ S implies S abelian (S ∩ F = {1} because p does not divide |F|).

• iv) If |F|= p ⁿ, |Aut(F)∣= φ(p ⁿ) = p ⁿ − p ⁿ⁻¹ = p ⁿ⁻¹(p − 1). Since G/F is isomorphic to a subgroup of Aut(F), if q ≠ p is a prime divisor of |G|, then q divides |G/F| and so also |Aut(F)| = p ⁿ⁻¹(p − 1). Hence q divides p − 1, so q < p. ⋄

The theorem we now prove may be viewed as a generalization of Sylow’s theorem in the case of solvable groups. If |G| = p ⁿ m, where p∤m, Sylow’s theorem ensures the existence of a subgroup of order p ⁿ. However, if we consider a different decomposition of the order of G into relatively prime integers, |G| = ab, with (a, b) = 1, nothing can be said, in general, about the existence of a subgroup of order a. For instance, the group A ₅ has order 60 = 15 · 4, (15, 4) = 1, but A ₅, has no subgroups of order 15. However, if the group is solvable, then such a subgroup always exists, and there also suitable generalization of the other parts of Sylow theorem.

Theorem 5.49 (Ph. Hall).

Let G be a finite solvable group of order |G| = ab, where (a, b) = 1. Then:

• i) there exists in G a subgroup of order a;

• ii) any two subgroups of order a are conjugate;

• iii) if A′ is a subgroup whose order divides a, then A′ is contained in a subgroup of order a.

Proof. By induction on the order of G. Let N be a minimal normal subgroup of G, hence a p-group: |N| = p ^k for some prime p. We shall prove i) and ii) simultaneously by distinguishing two cases, according as p|a or p|b.

• 1. p|a. Consider G/N. Then:

  

  By induction, there exists A/N ≤ G/N of order |A/N| = a/p ^k; hence |A| = a, and we have i). As for ii), the order of A being coprime to the index and p||A∣, the highest power of p dividing |G| also divides |A|. In other words, A contains a p-Sylow S of G. If A ₁ is another subgroup of order a, then for the same reason it contains a p-Sylow of G, S ₁ say. N being normal, N is contained in all the p-Sylows, and in particular in S and S₁, and so also in A and A ₁. Then A/N and A ₁/N are two subgroups of G/N of order a/p ^k, and therefore they are conjugate in G/N. It follows that A and A ₁ are conjugate in G, and this proves ii).

• 2. p|b. In this case,

  

  and always by induction there exists in G/N a subgroup L/N of order a. Then |L| = a · p ^k. If L < G, again by induction there exists in L (and so in G) a subgroup of order a. Now let L = G; we have:

  

  so N is Sylow in G. Consider now a minimal normal subgroup K/N of G/N. Then ∣K/N∣ = q ^t, where q is a prime different from p because N being a p-Sylow, p ∤ |G/N|. Hence |K| = q ^t p ^k and K = QN, where Q is a Sylow q-subgroup of K. Now K ⊴ G, so the Frattini argument implies

  

  Hence,

  

  from which a = |N _G (Q)|/|N ∩ N _G (Q)|, and therefore a divides |N _G (Q)|. If N _G (Q) < G, the existence of a subgroup of order a in N _G (Q), and so in G, is obtained by induction. If N _G (Q) = G, then G contains a normal q-subgroup whose order divides a, and we proceed as in case 1, with p replaced by q. The existence in case 2 is proved.

As to conjugation, let A and A ₁ be of order a; then AN/N and A ₁ N/N are also of order a (A ∩ N = {1} because p ∤ a) and therefore are conjugate in G/N. It follows that AN and A ₁ N are conjugate in G : AN = (A ₁ N)^g = A ^q₁ N, and so A, A ^g₁  ⊆ AN. If AN < G, since ∣AN∣ = a · pk, by induction A and A ^g₁ are conjugate in AN, A = A ^gx₁ , x ∈ AN, and so in G. Then let AN = G. N is Sylow and normal, and we may assume that it is the unique minimal normal subgroup (if there is another one, it cannot be a p-group for the same p of |N|, otherwise it would be contained in N, because N is the unique p-Sylow, contradicting the minimality of N; hence it is a q-group, q ≠ p, and therefore its order divides a and we are in case 1).

Let K/N be minimal normal in G/N. Then K/N is a q-group, with q ≠ p (N is Sylow, so p ∤|K/N|), and therefore K = NQ, where Q is a q-Sylow of K. Since K ⊴ G, the Frattini argument yields G = N · N _G (Q). We will show that N _G (Q) has order a, and that any subgroup of order a is conjugate to it. Set N _G (Q) = H, and consider N ∩ H; this subgroup centralizes Q (N ∩ H which is normal in H because N ⊴ G, and Q ⊴ H, have trivial intersection and are normal in H) and also N (because N is abelian). Hence, N ∩ H ⊆ Z(K), and Z(K) ⊴ G because Z(K) is characteristic in K ⊴ G. If N ∩ H ≠ {1}, then Z(K) ≠ {1}, so the latter subgroup contains a minimal normal subgroup, that is N, which is the unique such. Hence N centralizes K and therefore Q, so G = N.H = H, Q ⊴ G, N ⊴ Q, which is absurd. Hence N ∩ H = {1}, and G = N .H implies |G| = |N | · |H|, so |H| = a.

Let A be of order a, and consider AK. We have AK = G = AH, |AK/K | = |G/K| = ap ^k/|Q|p ^k = a/|Q|, and from |AK/K| = |A/(A ∩ K)| = a/|Q| it follows |A ∩ K| = |Q|. Therefore Q ₁ = A ∩ K is Sylow in K and hence is conjugate to Q in K. Moreover, from K ⊴ G it follow’s A ∩ K ⊴ A, so A ⊆ N _G (Q ₁) = H ₁. But Q ₁ ~ Q implies H ₁ ~ H, and therefore |H ₁| = |H| = a. Then A = H ₁ and A ~ H.

We now pass to the proof of iii). As above, we distinguish two cases.

• 1. p ^k|a. Here |A′N| divides a; otherwise there is a prime q that divides |A′N| and b, and so also either |A′| and b, or |N| and b, both cases being excluded. It follows that |A′N/N| divides a/p ^k; by induction, A′N/N ⊆ A/N, where |A/N| = a/p ^k. Then A′N ⊆ A, so A′ ⊆ A, where |A| = a.

• 2. p ^k|b. In this case, A′ ∩ N = {1} so that |A′N/N| = |A′|, and therefore, by induction, A′N/N ⊆ B/N, with |B/N| = a. Hence A′N ⊆ B, and |B| = a · p ^k. If B < G, by induction A′ is contained in a subgroup of B, and therefore of G, of order a. So let B = G. The order of G is now |G| = a · p ^k, and by i) there exists in G a subgroup A of order a. It follows G = AN, and a fortiori G = A · NA′. Hence:

  

  from which |A ∩ NA′| = |A′|, because |NA′| = p ^k|A′| since p ∤ a. The two subgroups A ∩ NA′ and A′ have order |A′|, they are contained in NA′, and their order being coprime to the index in NA′ by ii) they are conjugate there: A′ = (NA′ ∩ A)^g, with g ∈ NA′. Then A′ = NA′ ∩ A ^g, from which A′ ⊆ A ^g, and A ^g is the required subgroup. ◊

If the group is not solvable, there are counterexamples to all three parts of the theorem. We have seen that the group A ⁵, of order 60 = 4 · 15, has no subgroups of order 15. Also, 60 = 12 · 5, and A ⁵ contains subgroups of order 6 (for instance the group S ³ generated by (1, 2, 3) and (1,2)(4,5)) not contained in a subgroup of order 12 (these are all of type A ⁴). Hence i) and iii) do not hold. As for ii), in the simple group GL(3, 2) of order 168 = 24 · 7, there are subgroups of order 24 that are not conjugate (cf. 3.8.1, viii)).

Definition 5.17.

A subgroup of order coprime to its index is called a Hall subgroup.

If H is a Hall subgroup of a group G of index a power of a prime p (and therefore necessarily the highest power of p dividing |G|), and S is a Sylow p-subgroup, then S ∩ H = {1} and G = SH. Hence H is a p-complement of G (cf. Definition 5.7). With this terminology, i) of Hall’s theorem may be expressed by saying that a solvable group has a p-complement for all p ⁹. Unlike normal p-complements, in a solvable group a p-complement is not in general unique (think of the 2-complements of S ³).

Definition 5.18.

A Sylow basis for a finite group is a family of pairwise commuting Sylow p-subgroups, one for each prime dividing the order of the group.

Theorem 5.50.

In a finite solvable group there always exists a Sylow basis.

Proof. Let $\left|G\right|={\displaystyle {\prod}_{i=1}^t{p}_i^{k_i},}$ and let K _i be p _i -complements, one for each p _i . Then the K _i are of relatively prime indices, so the index of the intersection of any number of them is the product of the indices (Corollary 2.1). Consider ∩ _i≠j K _i ; the index of this subgroup is the product of all the ${p}_i^{k_i}$ except ${p}_j^{k_j},$ so its order is ${p}_j^{k_j},$ and therefore is a Sylow p _j -subgroup. The Sylow obtained in this way form a Sylow basis. Indeed, let Sj =|∩ _i≠j K _i and S _h = ∩ _i≠h K _i ; we show that S _j , S _h = S _h S _j . First, S _j = ∩ _i≠j K _i ⊆ ∩ _i≠j K _i , and the same for S _h . Then S _j , S _h ⊆ ∩ _i≠j,h K _i ; this intersection is a subgroup of order ${p}_j^{k_j}{p}_h^{k_h},$ and since it contains the set S _j S _h , that has the same number of elements, they are equal. Then S _j S _h is a subgroup, and therefore S _j S _h = S _h S _j . ◊

By the theorem above, a finite solvable group is the product of pairwise commuting nilpotent groups.

Theorem 5.49 shows how in a solvable group the Hall subgroups have properties similar to those of the Sylow subgroups. In the case of a Sylow S of a group G we know that N _G (N _G (S)) = N _G (S), and we have proved this result as an application of the Frattini argument. A suitable generalization of this argument will yield the same result for the Hall subgroups of a solvable group.

Theorem 5.51 (Generalized Frattini Argument).

Let K ₁ and K ₂ be subgroups of a normal subgroup H of a finite group G with the property that if they are conjugate in G, then they are already conjugate in H. Then:

Proof. Let g ∈ G. Then K ^g₁  = K ₂ is contained in H, so there exists h ∈ H such that k ₁ = k ^gh₁ . It follows gh ∈ N _G (K ₁) and the result. ◊

If H ⊴ G and K ₁ is Hall in H, then k ₂ = k ^g₁ is contained in H and is Hall in it. By ii) of the Hall theorem applied to H, K ₁ and K ₂ are conjugate in H, and the assumptions of Theorem 5.51 are met.

Corollary 5.15.

Let H be a Hall subgroup of a solvable group G. Then:

Proof. Setting K ₁ = H, H = N _G (H) and G = N _G (N _G (H)) in Theorem 5.51 gives the result. ◊

Definition 5.19.

A subgroup of a group is a Carter subgroup if it is nilpotent and self-normalizing.

A Carter subgroup H is a maximal nilpotent subgroup. Indeed, if H < K and K is nilpotent, then H < N _k (H), against H being self-normalizing.

Theorem 5.52 (Carter).

A finite solvable group always contains Carter subgroups, and any two such subgroups are conjugate.

Proof. We first prove existence. Let N be a minimal normal subgroup of G, |N| = p ^k. By induction, G/N contains a Carter subgroup K/N. By nilpotence, K/N = S/N × Q′/N, where S/N is a Sylow p-subgroup of K/N and Q′/N is Hall. Then K = SQ′ and Q′ = QN, where Q is Hall. From Q′/N ⊲ K/N it follows Q′ = QN ⊲ K, and Q being Hall in QN the generalized Frattini argument yields K = QNN _K (Q) = NN _K (Q). The subgroup H = N _K (Q) is a Carter subgroup. Indeed:

• 1. H is nilpotent.

  We have K = SQ′ = SNQ = SQ, so H = H ∩ K = Q · (H ∩ S) (Dedekind’s identity, ex. 19 with A = Q, B = S and C = H). Moreover, S ⊲ K because S/N ⊲ K/N, and therefore H ∩ S ⊲ H · Q is nilpotent because Q′/N is nilpotent: indeed, for this quotient we have Q′/N = QN/N ≃ Q/(Q ∩ N) ≃ Q, because Q ∩ N = {1}. Since Q ∩ (H ∩ S) = {1}, the two subgroups commute elementwise H = Q × (H ∩ S), so H is the direct product of two nilpotent subgroups, and as such is nilpotent.

• 2. N _G (H) = H.

  Let g ∈ G be such that H ^g = H. From K = NH and N ⊲ G, we have K ^g = N ^g H ^g = NH = K. Then (K/N)^gN = K/N, and since K/N is Carter in G/N it is self-normalizing in G/N. Hence g ^N ∈ K/N, and g ∈ K. But in K the subgroup Q is Hall, and therefore its normalizer does not grow: N_K(H) = H, and g ∈ H.

Let us now prove that two Carter subgroups are conjugate. Assume by induction that this holds for a solvable group of order less that |G|, and under this inductive hypothesis let us show that if H is Carter in G and N ⊴ G, then the image HN/N of H is Carter in G/N. Nilpotence is obvious. Let (HN/N)^gN = HN/N. Then (HN)^g = HN, i.e. H ^g N = HN. If g ∉ HN, then HN < G, and the two Carter subgroups H ^g and H of HN are conjugate in HN by induction: (H ^g)^y = H, y ∈ HN. It follows gy ∈ N _G (H) = H, g ∈ Hy ⁻¹ ⊆ H · HN = HN. Hence we have the contradiction that from g ∉ HN follows g ∈ HN. Therefore g ∈ HN, gN ∈ HN/N, and HN/N is self-normalizing.

Always under the above inductive hypothesis, let H ₁ and H ₂ be two Carter subgroups of G. Then H ₁ N/N and H ₂ N/N are Carter in G/N, and therefore by induction they are conjugate in G/N. Then H ₁ N and H ₂ N are conjugate in G : (H ₁ N)^g = H ₂ N, H ^g₁ N = H ₂ N, and if this subgroup is not the whole of G, by induction H ^g₁ and H ₂ are conjugate in it, and therefore in G. Hence we may assume G = H ₁ N = H ₂ N, and N abelian (by taking N minimal normal). Then H1 ∩ N ⊴ N (because N is abelian), H ₁ ∩ N ⊴ H ₁ (because N ⊴ G, and therefore H ₁ ∩ N ⊴ G). By the minimality of N, H ₁ ∩ N = N or H ₁ ∩ N = {1}. If H ₁ ∩ N = N, then N ⊆ H ₁, so G = H ₁ and there is nothing to prove. Then let H ₁ ∩ N = {1}, and similarly H ₂ ∩ N = {1}.

Furthermore, H ₁ and H ₂ are maximal subgroups. Indeed, if H ₁ ⊆ M ⊆ G, then M ∩ N ⊴ M (because N is normal) and M ∩ N ⊴ N (because N is abelian), and therefore M ∩ N ⊴ G because G = H ₁ N = MN. By the minimality of N, either M ∩ N = {1} or M ∩ N = N .In the former case, G/N = MN/N ≃ M and G/N = H ₁ N/N ≃ H ₁ (if M ∩ N = {1}, then a fortiori H ₁ ∩ N = {1}), and M = H ₁; in the latter, N ⊆ M and M = G.

Finally, let Q ₁ and Q ₂ be two p-complements of H ₁ and H ₂, respectively. Then they are two p-complements of G and therefore by Hall’s theorem they are conjugate: Q ₁ = g ⁻¹ Q ₂ g. Let us prove that g conjugates H ₂ and H ₁. Indeed, let H ₁ = g ⁻¹ H ₂ g; then Q ₁ ⊆ H ₁, Q ₁ ⊆ g ⁻¹ H ₂ g, and Q ₁ is Hall in these two subgroups, and therefore normal by nilpotence. Hence it is normal in the subgroup they generate, which is G because H ₁ is maximal. The factor group G/Q ₁ is a p-group (Q ₁ is a p-complement) and properly contains H ₁/Q ₁, which is Carter in G/Q ₁, and hence self-normalizing; but this is impossible in a p-group. This contradiction shows that H ₁ and H ₂ are conjugate. ◊

In nonsolvable groups Carter subgroups may not exist. In the simple group A ₅ the only nilpotent subgroups are the Sylow subgroups and their subgroups. But N _G (C ₂) is a Klein group, N _G (V) = A ₄, NG(C ₃) = S ₃ and N _G (C ₅) = D ₅.

Remark 5.6.

The notion of a nilpotent self-normalizing subgroup corresponds to that of a Cartan subalgebra of a Lie algebra. We give a brief explanation of this relationship. An algebra is a ring that is also a vector space. More precisely, let K be a field and let A be a ring with a vector space structure on the additive group of A such that α(ab) = (aα)b = a(αb), α ∈ K, a,b ∈ A. This algebra is a Lie algebra ℒ if in addition:

• i) ab + ba = 0;

• ii) (ab)c + (bc)a + (ca)b = 0

(equality ii) is the Jacobi identity). Then the product ab is denoted [a,b] (Lie product). A subalgebra $\mathcal{B}$ of a Lie algebra is a subspace which is closed under the Lie product, i.e. if $a,b\in \mathcal{B},$ then $\left[a,b\right]\in \mathcal{B}.$ An ideal of a Lie algebra is a subspace I such that if x ∈ I, then [a,x] ∈ I for each a ∈ ℒ (since, by i), [a,b] = −[b,a], there is no distinction between right and left ideals). Starting from an associative algebra one obtains a Lie algebra by setting [a, b] = ab — ba. This is the case, for instance, of the matrix algebra over a field. Subalgebras are defined in the obvious way. Let $\mathcal{B}$ be a subalgebra of ℒ. Denote ${\mathcal{B}}^{\prime }=\left[\mathcal{B},\mathcal{B}\right],$ the subalgebra generated by all products [b1,b2] as b1 and b2 vary in $\mathcal{B}$, that is the smallest subalgebra of ℒ containing all these products, and define inductively ${\mathcal{B}}^{(k)}=\left[{\mathcal{B}}^{\left(k-1\right)},\mathcal{B}\right]\left({\mathcal{B}}^{(0)}=\mathcal{B}\right).$ We have ${\mathcal{B}}^{(k)}\supseteq {\mathcal{B}}^{\left(k-1\right)},$ and if ${\mathcal{B}}^{(n)}=\left\{0\right\}$ for some n the subalgebra is nilpotent (abelian, if n = 1). The normalizer $\mathrm{N}\left(\mathcal{B}\right)$ of the subalgebra $\mathcal{B}$ is the set $\mathrm{N}\left(\mathcal{B}\right)=\left\{a\in \mathcal{L}|\left[a,b\right]\in \mathcal{B},\forall b\in \mathcal{B}\right\},$ which is a subalgebra (as follows from Jacobi identity). It is the largest subalgebra of ℒ containing $\mathcal{B}$ as an ideal (as in the case of groups, where the normalizer is the largest subgroup in which a subgroup is normal). A nilpotent selfnormalizing subalgebra $\mathcal{B},\mathbf{N}\left(\mathcal{B}\right)=\mathcal{B},$ is called a Cartan subalgebra. An example of a Cartan subalgebra in the matrix algebra (with the Lie product [A, B] = AB — BA, where AB is the usual matrix product) is the subalgebra $\mathcal{B}$ of diagonal matrices. It is a subalgebra, as is easily seen. Moreover, [A, B] = 0, because two diagonal matrices commute, so ${\mathcal{B}}^{\prime }$ = 0, and $\mathcal{B}$ is nilpotent (and in fact abelian). Let us show that it is selfnormalizing. Let $\left[A,B\right]\in \mathcal{B},$ for all $B\in \mathcal{B}$, that is, let [A, B] be a diagonal matrix if B is. Then AB – BA is diagonal; however, this matrix has all zeroes on the main diagonal, and since it is a diagonal matrix, it is the zero matrix. It follows AB = BA, and since a matrix commuting with all the diagonal matrices is itself diagonal, we have $A\in \mathcal{B}$.

We close this chapter with a sufficient condition for the solvability of a finite group due to I. N. Herstein. First a lemma.

Lemma 5.10.

In Theorem 5.39 assume further that H is abelian. Then:

• i) the restriction to H of the transfer V : G ⟶ H is the identity on H, that is V(h) = h, for all h ∈ H;

• ii) G = HK, K = ker(V) and by i), H ∩ K = {1}.

Proof. i) We know that $V(h)={\displaystyle {\prod}_{i=1}^t{x}_i{h}^{r_i}{x}_i^{-1},}$ where ${x}_i{h}^{r_i}{x}_i^{-1}\in H.\mathrm{If}\;{x}_i\ne \left\{1\right\},$ then ${h}^{r_i}\in H\cap {H}^{x_i^{-1}}=\left\{1\right\}.$ In the product, the only contribution is that of x1 = 1, i.e. $V(h)={h}^{r_1}.$ By (5.13) with x1 = 1, the cycle of σ to which 1 belongs contains only 1; indeed, 1 ∙ h = h1xσ(1), from which x _σ(1) = h ^− 1₁ h ∈ H, so xσ(1) = 1, σ(1) = 1 and r1 = 1. It follows V(h) = h. ii) By i), V is surjective (H is already surjective), so H ≃ G/K, K = ker(V), from which |G| = |H| ⋅ |K| = |HK| because H ∩ K = {1} (since by i) V(h) = 1 implies h = 1). ◊

If H is not abelian, this lemma yields V(G) = H/H′, and the restriction of V to H is the canonical homomorphism h → hH′.

If in a Frobenius group the stabilizer of a points is abelian, then by the lemma above G _α = H has |K| conjugates, where K is the kernel of the transfer of G in H; therefore, there are (|H| – 1)|K| = |G| – |K| elements that move points. Hence, those moving no points are |K| in number, and are the elements of the subgroup K

Theorem 5.53 (Herstein).

Let G be a finite group admitting an abelian maximal subgroup H. Then G is solvable.

Proof.¹⁰ If N _G (H) > H, then N _G (H) = G by the maximality of H; therefore H ⊴ G and G/H is of prime order (always because of the maximality of H). Hence H and G/H are both solvable, and so also is G. Then let N _G (H) = H, and let B = H ∩ H ^x ≠ {1} for x ∉ H. Then B is normal in both H and H ^x (these are abelian groups), so N _G (B) properly contains H; hence B ⊴ G. G/B contains the abelian maximal subgroup H/B; by induction on |G|, G/B is solvable, and B being solvable, so also is G. Hence we may assume H ∩ H ^x = {1} for x ∉ H. Let K be the kernel of the transfer of G in H, and for a given p let S be the Sylow p-subgroup of K fixed by conjugation by the elements of H (es. 54; 1 ≠ h ∉ H acts f.p.f. on K). Then HS is a subgroup properly containing H, so HS = G, and since H normalizes S we have S ⊴ G (from HK = G we have S = K). S being nilpotent, and therefore solvable, and G/S ≃ H being abelian, G is solvable. ◊

Remark 5.7.

If the hypothesis “H abelian” is replaced by “H nilpotent”, then Theorem 5.53 is not true in general. A deep result of J.G. Thompson allows one to establish solvability if the group admits a maximal nilpotent subgroup of odd order.

Theorem 5.54 (Schmidt-Iwasawa).

A finite group in which every proper subgroup is nilpotent is solvable.

Proof. Let G be a minimal counterexample. i) G is simple. Deny, and let {1} < N ⊲ G. N is nilpotent; if H/N < G/N, H is nilpotent and so also is H/N; by induction, G/N is solvable, and N being nilpotent, G is solvable, against the choice of G. ii) If L and M are two distinct maximal subgroups of G, then L ∩ M = {1}. Deny, and among all pairs of maximal subgroups let L and M be such that L ∩ M = I is of maximal order. We have I < L and I < M (because L ≠ M), and by nilpotence I <N _L (I) and I <N _M (I). Moreover N _G (I) < G, because G is simple, and therefore there exists a maximal subgroup H that contains N _G (I); it follows L ∩ H ⊇ L ∩ N _G (I) = N _L (I) > I, from which |L ∩ H| > |I| = |L ∩ M|. The choice of L and M implies H = L; similarly, H = M, and therefore L = M, whereas L ≠ M. iii) A group G satisfying i) and ii) does not exist. Deny; then every element of G belongs to a maximal subgroup, and by ii) to only one. By i), none of these is normal, hence by maximality they are self-normalizing. Then the group is a union of subgroups pairwise with trivial intersection and self-normalizing. But such a group cannot exist (Chapter 2, ex. 39).

This theorem applies in particular to the study of nonnilpotent groups in which every subgroup is nilpotent. Such a group is called a minimal nonnilpotent group.

Corollary 5.16.

The order of a minimal nonnilpotent finite group G is divisible by exactly two primes.

Proof. By the theorem above G is solvable, hence it admits a normal subgroup N of prime index p. If Q is a Sylow q-subgroup q ≠ p, then Q ⊂ N and is Sylow in N, therefore characteristic in N and so normal in G. If PQ < G for all Q, then these subgroups PQ are nilpotent, so P is normalized by all the Sylow q-subgroups, and so is normal in G. We already know that the Sylow q-subgroups are normal, so G would be nilpotent, against the assumption. It follows PQ = G, for some Q. ◊

5.10 Exercises

59.

• i) The nonsimple groups of ex. 38, 41, 43 e 44 of Chapter 3 (except those of order 180 and 900) are solvable.

• ii) the dihedral groups D _n are solvable, for all n.

60. The inverse of Lagrange’s theorem does not hold in general for solvable groups.

61.

• i) The product of two solvable groups, with one of which normal, is solvable (cf. Theorem 5.11);

• ii) the direct product of a finite number of solvable groups is solvable.

62. If a group G admits an automorphism sending an element to itself or to its inverse, then G is solvable. [Hint: if H is the subgroup of the elements fixed by the automorphism, then H is normal and abelian, and G/H is also abelian.]

63. Let G be a finite solvable group and let σ be an automorphism fixing every element of the Fitting subgroup F of G. Prove that (o(σ), |G|) ≠ 1. [Hint: given g ∈ G, let g ^σ = gx; letting g act on F by conjugation, prove that x centralizes F, hence x ∈ F and therefore is fixed by σ. Moreover, ${g}^{\sigma^k}=g{x}^k,$ for all k, and o(x)|o(σ).]

64. Prove the equivalence of the following propositions:

• i) a group of odd order is solvable¹¹;

• ii) a nonabelian simple group has even order.

65. Let G be solvable, H ⊴ G. Then HG′ < G (cf. ex. 6).

66. Let G be a solvable finite group in which every Sylow subgroup coincides with its normalizer. Prove that G is a p-group.¹² [Hint: let H be a normal subgroup of index a prime p. Then H contains a q-Sylow Q; apply the Frattini argument.]

67. Let G be a solvable primitive permutation group of degree n. Prove that n is a power of a prime.

68. A maximal subgroup in a finite solvable group has prime power index (cf. Corollary 5.2).

69. Let H and K be two maximal subgroups of relatively prime order of a finite solvable group. Prove that their indices are also relatively prime. The converse is not true.

70. Let G be a finite solvable group in which every maximal subgroup has order coprime to the index. Prove that for some p a p-Sylow is normal.

71. Let G be a finite solvable group. Prove that if x ₁ x ₂ ⋯ x _n = 1 and the x _i are of coprime order, then x _i = 1 for all i.¹³

72. Let G be a finite solvable group, and let S ₁, S ₂, …, S _t be Sylow subgroups, one for each prime dividing |G|. Prove that the product of the S _i , in any order, is the whole group G (cf. Ex. 3.4, 12).

73. Let G be a group. If two consecutive factor groups of the derived series G ^i–1/G ⁱ and G ⁱ/G ⁱ⁺¹, i > 1 are cyclic, then G ⁱ = G ⁱ⁺¹. [Hint: prove that setting H = G ^i–2/G ⁱ⁺¹, the factor group H′/Z(H′) is cyclic, and H′ abelian.]

74. If G has all its Sylow cyclic, then:

• i) G is solvable;

• ii) G′ and G/G′ are cyclic (such a group is called metacyclic).

75. A maximal solvable subgroup of a finite group is self-normalizing.

76. (Galois) A transitive subgroup G of S ^p, p a prime, is solvable if and only if it contains a normal subgroup of order p. In other words, G is solvable if and only if its action is affine (Chapter 3, ex. 67).

Remark 5.8. In Galois theory, the result of the previous exercise has the following meaning: an irreducible equation of prime degree is solvable by radicals if and only if its roots may be expressed as rational function of any two of them. Let us see why.¹⁴ If the Galois group is solvable, then being a subgroup of S ^p, by the above is an affine group, so if an element fixes two points it is the identity. Adjoining two roots α _i and α _j to the field, the Galois group reduces to a subgroup that fixes these two roots, and so is the identity. Hence K(α ₁, α ₂, …, α _p ) = K(α _i , α _j ), so every root is a rational function of α _i and α _j . Conversely, if the roots are all rational functions of two of them, α _i and α _j , say, then K(α ₁, α ₂, …, α _p ) = K(α _i , α _j ); this equality means that if an element of the Galois group fixes two roots it fixes all of them, and therefore is the identity. Hence the group is an affine group, and therefore solvable.

77. Prove that, under the hypothesis of Theorem 5.53, G‴ = {1}.

78. Let p ₁, p ₂, …, p _n be the composition factors of a solvable group G (possibly with repetitions). Then G is nilpotent if and only if for every permutation ${p}_{i_1},{p}_{i_2},\dots, {p}_{i_n}.$ of the p _i , G admits a composition series whose composition factors appear in the order ${p}_{i_1},{p}_{i_2},\dots, {p}_{i_n}.$

79. A finitely generated periodic solvable group is finite (cf. Theorem 5.17).

80. Let G be finite, G = ABA, where A is an abelian subgroup and B is of prime order. Prove that G is solvable. [Hint: use Theorem 5.53.]

81. Prove that:

• i) a subgroup A of a group G is a maximal abelian subgroup if and only if it equals its centralizer;

• ii) if the Fitting subgroup F of G is abelian, then it is the only maximal abelian normal subgroup;

• iii) if F is abelian, and G is solvable, then F is a maximal abelian subgroup.

82. Let G be a finite group in which every maximal subgroup has prime index. Prove that if q is the largest prime dividing |G|, then a Sylow q-subgroup is normal, and G is solvable. [Hint: if Q is not normal, let M be maximal, M ⊇ N _G (Q); consider the number of Sylow q-subgroups of G and M.]

83.

• i) Let G be a simple group, G = HK, where H and K are two proper subgroups with H abelian. Prove that H ∩ K = {1}.

• ii) Let |G| = p ^a q ^b and let the Sylows be abelian. Prove that the group is solvable. [Hint: it is sufficient to prove that G is not simple (use i)) and induct.] We shall see in the next chapter that the result also holds without the assumption of the Sylows being abelian (Theorem 6.20).

84. A free group of rank greater than 1 is not solvable.

85. The group G of Ex. 4.1, 4 is solvable (Chapter 4, ex. 12), but not polycyclic (Theorem 5.21).

Definition 5.20. A group is supersolvable if it has an invariant series with cyclic quotients. (In the definition of a polycyclic group replace “normal” with “invariant”.)

86. Subgroups and factor groups of supersolvable groups are supersolvable.

87. (Wendt) The derived group G′ of a supersolvable group is nilpotent. [Hint: let H/K be a quotient of the series of G′ obtained by intersecting G′ with a series of G with cyclic quotients. Two elements of G′ induce on H/K commuting automorphism, so their commutator induces the identity, and the series of G′ is central.]

88. Let G be a supersolvable finite group, and let $\left|G\right|={p}_1^{h_1}{p}_2^{h_2}\cdots {p}_t^{h_t}$ where p ₁ > p ₂ > … > p _t , and for each p _i let S _i be a Sylow p _i -subgroup. Prove that G contains a set of normal subgroups

where the S _i are Sylow p _i -subgroups, as follows:

• i) G has a normal subgroup N of prime order;

• ii) by induction, G/N admits a normal Sylow p-subgroup SN/N where p is the largest prime divisor of |G/N|;

• iii) if p = p ₁, then S = S ₁ is normal in G;

• iv) if p ≠ p ₁, then S is the unique p ₁-Sylow of SN, so S = S ₁ is characteristic in SN and therefore normal in G;

• v) by induction, G/S contains the normal subgroups S ₂, S ₂ S ₃, …, S ₂ S ₃ ⋯ S _t .

(A set of normal subgroups as in (5.15) is called a Sylow tower.)

89.

• i) The symmetric group S ⁴ is solvable but not supersolvable;

• ii) the derived group of S ⁴ is not nilpotent;

• iii) S ⁴ has a normal subgroup N such that both N and S ⁴/N are super-solvable.

90.

• i) A maximal subgroup of a supersolvable group has prime index. [Hint: use i) of ex. 88.]

• ii) Let G be a finite group in which the maximal subgroups are of prime index. Prove that, if q is the largest prime dividing |G|, a q-Sylow is normal and G is solvable. [Hint: if Q is not normal, let M a maximal subgroup containing N _G (Q); consider the number of q-Sylow of G and of M.] (By a theorem of Huppert, a finite group in which the maximal subgroups are of prime index is supersolvable.)