The $\mathcal{X}$-series of a $p$-group and complements of abelian subgroups

Let $G$ be a $p$-group. We denote by $\mathcal{X}_i(G)$ the intersection of all subgroups of $G$ having index $p^i$, for $i \leq \log_p(|G|)$. In this paper, the newly introduced series $\{\mathcal{X}_i(G)\}_i$ is investigated and a number of results concerning its behaviour are proved. As an application of these results, we show that if an abelian subgroup $A$ of $G$ intersects each one of the subgroups $\mathcal{X}_i(G)$ at $\mathcal{X}_i(A)$, then $A$ has a complement in $G$. Conversely if an arbitrary subgroup $H$ of $G$ has a normal complement, then $\mathcal{X}_i(H) = \mathcal{X}_i(G) \cap H$.

In fact, the previous question was the exact point of departure for the work reported here. Although we have not been able to formulate such a condition, we have found a condition which is sufficient (and in some cases necessary). We delay presenting the said result to discuss properties of the X -series of a p-group first. This is done in Section 2. Briefly, given a p-group G, we define X i (G) to be the intersection of all subgroups of G having index p i in G for i log p (|G|) and (for technical reasons) extend its definition to all non-negative integers. We call the associated series {X i (G)} i the X -series of G. This series has several nice properties: • It is subgroup-monotone; cf. Proposition 2.4.
• For every i, j 0 we have X i (X j (G)) X i+j (G). Moreover, successive quotients of the X -series of G are elementary abelian p-groups; cf. Lemma 2.5.
In Section 3 we resume our discussion by focussing first on abelian groups and what is known in that setting as regards complementation. We then utilise the tools assembled in Section 2 to prove our main result: Theorem 1.2. Let A be an abelian subgroup of the p-group G and suppose that for all indices i 0. Then A has a complement in G.
The importance of this result and its relevance to other known theorems will be further elucidated in due course. A sort of partial converse of Theorem 1.2 holds and its proof is given in Section 2: Theorem 1.3. Let H be an arbitrary subgroup of the p-group G and assume that H has a normal complement in G. Then for all indices i 0.
1.1. Notation. We outline below some notational conventions that we will use throughout the paper.
(ii) Let G be a p-group and let i be a non-negative integer. We will write H i (G) = g p i : g ∈ G for the subgroup generated by the p i -powers of elements of G. Clearly, H 0 (G) = G. The index i = 1 is sometimes dropped, so that H 1 (G) = H(G). These subgroups, for various indices i, are referred to as the agemo subgroups of G. We remark that G p i is sometimes used instead of H i (G) in the literature.
(iii) We will have occasion to refer to the Frattini series of a group. For an arbitrary group, the Frattini series is defined by Φ 0 (G) = G and Φ i (G) = Φ(Φ i−1 (G)), where Φ(G) is the Frattini subgroup of G, i.e. the intersection of all maximal subgroups of G.

P ro p e rt i e s o f t h e X -s e r i e s
We begin in earnest by defining the X -series of a p-group.
Definition 2.1. Given a p-group G and a positive integer i log p |G| we write for the collection of subgroups of G of index p i in G. We also write that is, X i (G) is the intersection of all subgroups of G that have index p i in G for all such i. We call the associated series the X -series of G and by convention we write X 0 (G) = G. Finally, if j > log p |G| then X j (G) = 1 by assumption.
Some first properties following the definition of the X -series are collected in the following lemma.
Lemma 2.2. If G is a p-group then the subgroups X i (G) are characteristic subgroups of G and Proof. (i) Clearly the X i (G) are characteristic subgroups of G, for all i, and as P 1 (G) is exactly the set of maximal subgroups of G.
(ii) To see this observe that for every j ∈ [k] we have due to the fact that for every subgroup S of index p k in a p-group G there exists a subgroup T G of index p k−j that contains S.
(iii) Now this part follows from (ii) as (iv) We induce on i. For i = 1 we clearly have Φ(G) M for every maximal subgroup M of G.
Assuming it holds for i, we will prove it for i + 1. Let T ∈ P i+1 (G) and pick K ∈ P i (G) with T K. So T is a maximal subgroup of K and Φ(K) T . Furthermore, the inductive hypothesis implies that Φ i (G) K and therefore (v) The proof of this part follows directly from (iv) and the definition of X i (G).
(vi) Since N X i (G), it follows from the correspondence theorem that this clearly proves the claim, and the proof of the lemma is complete.
Remark 2.3. Firstly, observe that equality really is possible in (iii), since in Q 8 , for example, we have X 1 (Q 8 ) = X 2 (Q 8 ). Secondly, we comment on the behaviour of the X -series under quotients.
Assuming that G is a p-group, as usual, and that N is an arbitrary normal subgroup of G, we have X 1 (G/N ) = X 1 (G)N N since X 1 (G) = Φ(G) = G H(G) and each of G , H(G) behaves well under quotients. However, things change for larger-index terms of the X -series. It is not true, for instance, that X 2 (G/N ) = X 2 (G)N N for arbitrary G, N . An example where this fails to be true is the group of order 16 with presentation The subgroup N = a 2 b 2 is central of order 2 and affords the quotient G/N ∼ = Q 8 . To see this, observe that G/N is non-abelian (since, for example, abN = baN ) and a 2 N = b 2 N is the unique element of order 2 in G/N . Now X 2 (G) is trivial (see Figure 1) and thus X 2 (G)N N is trivial, but X 2 (G/N ) is not.
F i g u r e 1 . The poset of classes of subgroups of G.
Next, we prove a useful property of the X -series.
Proposition 2.4. Let G be a p-group and let M G.
Proof. Fix a subgroup M G and an index i 0. To prove the desired containment, it suffices to show that X i (M ) K for every K ∈ P i (G). The equality |M K| |K| = |M | |M ∩ K| and the fact that K has index p i in G together imply that the index of K ∩ M in M is at most p i . Thus there exists a subgroup T of K ∩ M whose index in M is exactly p i . Therefore As this holds for every K ∈ P i (G), the proposition follows.
We will refer to this property of the X -series as the subgroup-monotonicity of X . Moreover, the same will be said for a subgroup series that satisfies the conclusion of Proposition 2.4 (i.e. that it is subgroup-monotone).
Lemma 2.5. Let G be a p-group. Then In addition, X i (G)/ X i+1 (G) is an elementary abelian p-group for every non-negative integer i.
Proof. In view of Lemma 2.2 we have X i+j (G) = T ∈P j (G) X i (T ). Let Then X i (K) X i (T ) for every T ∈ P j (G) by Proposition 2.4. Hence Thus, the first part of the lemma holds, from which we deduce that X i (G)/ X i+1 (G) is an elementary abelian group seeing as This completes the induction and the proof of the lemma.
We now look at the situation where some normal subgroup N of G is complemented in G. Theorem 1.3 is essentially the second part of the following: Theorem 2.6. Let G be a p-group and let N be a normal subgroup of G. Suppose that N is complemented, by H say, in G. Then for all i 0 we have: Proof. (i) The leftmost inclusion is a consequence of the fact that the product is well-defined (since X i (N ) is characteristic in N and the latter is normal in G) and the subgroup-monotonicity of the X -series.
As regards the right-hand-side inclusion, assume first that i log p (|H|). Observe that for all T, S ∈ P i (H) Dedekind's lemma yields (2.2) By (2.1) and (2.2) we get N S ∩ N T = N (S ∩ T ) and repeatedly applying this fact gives us is valid for all non-negative integers i, completing the proof.
(ii) We have that where the first inclusion follows from the subgroup-monotonicity of the X -series, the second inclusion from part (i) and the penultimate equality by Dedekind's lemma. This double inclusion forces X i (H) = X i (G) ∩ H, as desired.
Corollary 2.7. Let G be a p-group and let N be a normal subgroup of G that is complemented in G by H, say. Then N ∩ X i (G) has a complement in X i (G) for all i 0 and, in fact, X i (H) is one such complement.
Proof. Fix an index i 0. We show that X i (H) is a complement for N ∩ X i (G) in X i (G) by observing the following: The first equality is yet another application of Dedekind's lemma and the second equality follows directly from Theorem 2.6 (i). The proof is complete.
Theorem 2.6 will also aid us in the proof of the next result.
Theorem 2.8. Given p-groups G and H, we have For the other direction, note that by Theorem 2.6 we have and (2.4) Combining (2.3) and (2.4) gives us where all equalities are applications of Dedekind's lemma. Since both inclusions are valid, the proof of the theorem is complete.
In the general case, let G = C i × C 2 × · · · × C t , where C j are cyclic subgroups of G. Then in view of Theorem 2.8 we get X i (G) = t j=1 X i (C j ). But for any cyclic subgroup C we have where the penultimate equality follows from the fact that the Frattini subgroup respects direct products and an easy induction argument (see [Gas53,Satz 6]).
The last part of the proposition follows directly from the corresponding relations that the agemo subgroups of an abelian group 1 satisfy: The proof of the proposition is now complete.
Proposition 2.10. Let C be a cyclic subgroup of G of order p n . Then the following are equivalent: (i) C ∩ X n (G) = 1; (ii) C has a complement in G; We assume that C ∩ X n (G) = 1 and wish to establish the existence of a complement for C in G. Indeed, if C ∩ T > 1 for all subgroups T of index p n in G, then C ∩ T Ω 1 (C) > 1, as Ω 1 (G) is the unique subgroup of order p in C. Therefore C ∩ X n (G) Ω 1 (C) > 1, and this is clearly a contradiction. It follows that there exists at least one subgroup T of index p n in G whose intersection with C is trivial. That subgroup T serves as the desired complement.
(ii) → (iii) Let T be, as in the previous paragraph, a complement for C in G. As C is a cyclic group, we see that is the unique subgroup of C of order p n−i , we deduce that C ∩ K i = X i (C), and the claim follows. Now, K i has index p i in G and therefore X i (G) K i . We conclude that and thus X i (G) ∩ C = X i (C), as desired.
(iii) → (i) This follows easily from the fact that C ∩ X i (G) = X i (C) for all i 0 upon taking i = n and observing that X n (C) = 1.
Our next result outlines which members of the lower central series of G are guaranteed to be contained in some term of the X -series of G. As we will see, our result is essentially best possible.
Proposition 2.11. If G is a p-group and i is a positive integer then and the index p i−1 + 1 is optimal.
Proof. Let H be an arbitrary subgroup of G of index p i in G. The regular action of G on the left cosets of H in G gives rise to a homomorphism φ : G → S p i whose kernel N is the core of H in G; that is N = g∈G H g H. 1 In fact, this relation among the agemo subgroups is satisfied more generally by regular p-groups; see Lemma 1.2.12 (ii) in [LGM02], for example.
Let L i be a Sylow p-subgroup of S p i . Then G/N is isomorphic to a subgroup of L i . Now L i is an iterated wreath product C p . . . C p i terms having order p p i −1 p−1 and class c = p i−1 (cf. [Kal46]). Hence γ p i−1 +1 (L i ) = 1, so γ p i−1 +1 (G/N ) = 1. But the lower central series behaves well with respect to quotients and therefore We conclude that γ p i−1 +1 (G) N H and as H was arbitrary we see that To see that the index p i−1 + 1 is optimal, observe that γ p i−1 +1 (L i ) = 1 and the index j = p i−1 + 1 is the least positive integer such that γ j (L i ) = 1 (since L i has class p i−1 ). In proof of the second claim of the proposition therefore, it will be sufficient to establish that X i (L i ) = 1 for all positive integers i. We argue by induction on i, noting that the base case i = 1 is clearly valid since L 1 has prime order p and thus its Frattini subgroup is trivial. Assume validity of the claim for i − 1. Note that L i = L i−1 C p and that the base of L i On the other hand, we have by Lemma 2.2 (ii) (with k = i and j = i − 1) where the first equality follows from Theorem 2.8 while the second equality is valid by the inductive hypothesis. Thus, X i (L i ) = 1, completing the induction and the proof. We assert that both terms in the right-hand-side of the equality above are contained in X 2 (G). That γ 3 (G) X 2 (G) follows from Proposition 2.11 with p = i = 2. Thus, it will suffice to prove that [G, H(G)] X 2 (G). According to Corollary 1.28 in [LGM02], we have when G is a 2-group. In addition, where the first containment is justified by the subgroup-monotonicity of the agemo subgroup (cf. Lemma 1.2.7 in [LGM02]) and the last by Lemma 2.2 (v). Hence the claim has been proved.
The content of Corollary 2.12 is no longer valid if p > 2. Indeed, for any odd prime p the Sylow p-subgroup C p C p of S p 2 satisfies X 2 (C p C p ) = 1 since it has a maximal subgroup which is elementary abelian (in particular, its base), but its Frattini subgroup is not central. To see why this last claim is true, assume that Φ(G) Z(G) and recall that by [Kal46] the class of C p C p is p. Thus, if Φ(G) Z(G) then Φ(G) = Z(G), since Z(G) has order p. On the other hand G Φ(G) = Z(G), forcing γ 3 (G) to be trivial, which is absurd (even for p = 3).

C o m p l e m e n t s a n d a b e l i a n s u b g ro u p s
If A is an abelian group then a condition which is both necessary and sufficient for a subgroup B of A to have a complement in A exists and is based on Prüfer's notion of a pure subgroup; see for example [Fuc15,Chap. 5]. We define this next.
Definition 3.1. Let A be an abelian group and let B be a subgroup of A. Then B is called a pure subgroup of A provided that if a n ∈ B for a ∈ A and n ∈ N then there exists b ∈ B such that a n = b n . Equivalently, B is pure in A if and only if B n = B ∩ A n for all n ∈ N.
For the convenience of the reader we present here an independent proof of the following result. Note that no originality is claimed here. The result itself should be well-known to abelian group theorists (and, in fact, holds more generally than we have stated), but perhaps not so much to group theorists in general. Proof. Suppose first that B is complemented in A and let C be a complement. Then A is the (internal) direct product A = B × C. Thus, for each a ∈ A there exist unique b ∈ B and c ∈ C such that a = bc. Therefore, if a n ∈ B for some n ∈ N then b n c n ∈ B thus c n ∈ B ∩ C = 1. So a n = b n and thus B is pure in A.
For the converse, suppose that B is pure in A and that B C A, where C/B is cyclic. Let n = |C : B| and write cB = C/B. As c n ∈ B, there exists b ∈ B such that c n = b n . Setting g := cb −1 , we have g n = 1. Since gB = cB, we see that n = o(gB) in C/B. It follows that C = B × g and thus B is a direct factor of C. Now we focus our attention on p-groups. In view of Gaschütz's theorem mentioned in Section 1, this is natural. In this setting, the previous theorem asserts the following: Let G be an abelian p-group and let A be a subgroup of G. Then A has a complement in G if and only if A n = A ∩ G n ( †) for all n ∈ N.
However, the only relevant exponents n in this case are the powers of p, thus ( †) is essentially equivalent to H i (A) = A ∩ H i (G) ( ‡) for all non-negative integers i. On the other hand, the H i subgroup of an abelian p-group is its X i subgroup, according to Proposition 2.9. Thus we have the following.

Corollary 3.3. Let G be a abelian p-group and let A be a subgroup of G. Then A has a complement in G if and only if
A ∩ X i (G) = X i (A) for all indices i 0.
We are now ready to prove that the condition above, which is both necessary and sufficient for complementation of a subgroup of an abelian p-group, is sufficient in general.
Proof of Theorem 1.2. We begin with an easy observation; it will suffice to establish the claim for all indices i such that p i is at most the exponent of the subgroup A. Moreover, notice that if A is cyclic, then the equivalence between (ii) and (iii) in Proposition 2.10 ensures that A is complemented in G. Now we argue by induction on |A|, noting that we have addressed the base case of the induction in the previous paragraph. We may also assume that A is not cyclic. So let C be a cyclic subgroup of maximal order in A and note that |C| = exp(A) = p r , say. By assumption, A∩X r (G) = X r (A) = 1. Then C ∩ X r (G) A ∩ X r (G) = 1, thus C ∩ X r (G) = 1 and the equivalence between (i) and (ii) in Proposition 2.10 suffices to guarantee that C has a complement, say H, in G. Since the X -series is subgroup-monotone, we have for each i t the following: We follow up with a few remarks on the content and proof of Theorem 1.2.
Remark 3.4. It should be noted that Theorem 1.2 works for a class of p-groups which is strictly larger than the class of abelian p-groups. For A to have a complement in G it is sufficient, as the proof of the theorem indicates, that A have the following property: for every H A there exists a normal (in H) cyclic subgroup 1 < C ¢ H that admits a complement in H, For example, the property above is enjoyed by the dihedral 2-groups D 2 n which are non-abelian for n 3.
On the other hand, Theorem 1.2 is not true unconditionally; some assumption about A must be made. Take  Remark 3.5. On the surface, it looks as though a literal interpretation of purity might be enough to guarantee the existence of a complement in Theorem 1.2. That is to say, under the assumptions of the theorem, perhaps it is true that an abelian subgroup A of G satisfying A ∩ H i (G) = H i (A) for all i 0 is necessarily complemented in G. However, this is not so; purity really must be stated in an equivalent form. To see why, consider an odd prime p and let G be a non-abelian p-group of exponent p. Then it is easy to see that the equality A ∩ H i (G) = H i (A) is valid for every (abelian) subgroup of G, but clearly not every abelian subgroup can be complemented; simply consider a subgroup of X 1 (G) = Φ(G) of order p.
Corollary 3.6. Let G be a p-group and let A be a central subgroup of G. Then A has a complement in G if and only if A ∩ X i (G) = X i (A) for all i 0.
Proof. Since A is central in G, it is abelian. Thus Theorem 1.2 implies that A is complemented in G. Conversely, suppose that A has a complement in G, say H. Then H is normalised by both A and H, thus G = A × H and A is, in fact, normally complemented in G. Now Theorem 1.3 implies that X i (A) = A ∩ X i (G), as desired.
In fact, looking carefully at the proof of Theorem 1.2 and extracting the abstract properties of the X -series used in the proof allows us to state a generalisation of Theorem 1.2. We need a definition first.
Definition 3.7. Given a p-group G we call the series a good series for G if the following three conditions are satisfied: (ii) B i (G) (and thus W i (G) as well) is subgroup-monotone; (iii) B i (G) (and thus W i (G) as well) respects direct products; (iv) If C is a cyclic group, then Ω 1 (C) B 1 (C) (and thus Ω 1 (C) W 1 (C) as well).
Clearly, X i (G) itself is a good series for G. Other examples of good series are the following: The generalisation asserted earlier now reads as follows.
Theorem 3.8. Let A be an abelian subgroup of the p-group G and suppose that W i is a good series for G so that A ∩ W i (G) = W i (A) for all integers i 1. Then A has a complement in G.
Since the proof is nearly identical to that of Theorem 1.2, we have elected to omit it.
We remark that the following result, although resembling Corollary 2.7, is ultimately independent, since no normality assumption is made.
Proposition 3.9. Let G be a p-group and A an abelian subgroup of G such that for all non-negative integers i. Then A ∩ X i (G) has a complement in X i (G) for all i 0.
Proof. By Theorem 1.2 we have already established the case i = 0, as X 0 (G) = G and A has a complement in G. According to the same theorem, the abelian group A ∩ X i (G) has a complement in X i (G) provided we can show that for every integer t 1.
One inclusion is straightforward, as the X -series is subgroup-monotone and therefore For the other inclusion, we first observe that where the last equality follows from Proposition 2.9. Hence So we have shown that equation (3.1) is valid for every t 1, and the proposition follows.
Corollary 3.10. Let G be a p-group and let A be a normal abelian subgroup of G with the property A ∩ X i (G) = X i (A) for all non-negative integers i. Then A has a complement in G and for every complement H the analogous property Proof. Since A ∩ X i (G) = X i (A) for all i 0, Theorem 1.2 implies that A is complemented in G. If H is any such complement for A, then the fact that H ∩ X i (G) = X i (H) holds for all i 0 is a consequence of Theorem 1.3. Moreover, Corollary 2.7 implies that A ∩ X i (G), which by assumption is X i (A), is complemented in X i (G) by X i (H) and this holds, of course, for all i 0.
We conclude our work by pointing out the limitations of Theorem 1.2.
First of all, by virtue of Proposition 2.10 it is the case that if A is cyclic then it is complemented in G if and only if A ∩ X i (G) = X i (A) for all i 0. Moreover, the condition works as "if and only if" in case A is central in G owing to Corollary 3.6.
On the other hand, observe that an abelian subgroup A of a p-group G can be complemented in G without satisfying the condition A ∩ X i (G) = X i (A) for all i 0. Concrete examples abound; one possibility is to take A to be an elementary abelian group containing the Frattini subgroup. If A is complemented in G (and this happens for many groups, e.g. D 8 with A either of the two non-cyclic maximal subgroups), where G is non-abelian, then it cannot possibly be the case that A ∩ X i (G) = X i (A) for all i 0. Things go awry as quickly as possible, i.e. for i = 1.
In fact, an elementary abelian subgroup A satisfies A ∩ X i (G) = X i (A) for all i 0 if and only if A ∩ Φ(G) = 1, thus Theorem 1.2 is only useful when A avoids the Frattini subgroup of G. However, in that case Theorem 1.2 tells us nothing new since Satz 7 in Gaschütz's paper [Gas53] guarantees that A has a complement in G regardless of whether G is a p-group or not, as long as G is finite.
What is perhaps not so obvious is the following.
Proposition 3.11. Let G be a p-group and let A be a normal subgroup of G such that A∩Φ(G) = 1. Then A has a normal complement in G and is thus a direct factor of G.
Proof. First of all, observe that we do not need to assume that A is abelian since it follows from Φ(A) A ∩ Φ(G) = 1 that A must, in fact, be elementary abelian. Now we argue by induction on |A|. The claim is evident if A = 1, so the induction begins and we assume that A > 1. Since A ∩ Φ(G) = 1 and A > 1, there exists a maximal subgroup M of G such that A is not contained in M . Write D := A ∩ M . In case D = 1, we are done because M is normal in G. So suppose that D > 1. Observe that since AM > M and M is maximal in G, we have G = AM . Now both M and A normalise D, so D is normal in G.
The induction hypothesis applied to D in G ensures the existence of a normal complement K. In other words, G = D × K. By Dedekind's lemma, we have A = D × (A ∩ K). The subgroup A ∩ K is normal in K and proper in A. Also, Φ(G) = Φ(D) × Φ(K) = Φ(K). This holds because A is elementary abelian hence so is D.

Now
(A ∩ K) ∩ Φ(K) = A ∩ Φ(K) = A ∩ Φ(G) = 1. Applying the induction hypothesis to A ∩ K in K, we get K = (A ∩ K) × H for some subgroup H. But now H is normalised by K and also by D, because D actually centralises it, and since G = DK we see that H is normal in G. It is easy to see that A ∩ H = 1, so H is the required normal complement and that completes our induction.
It is clear that the same proof would work just as well if we had assumed more generally that G is a nilpotent group instead of a p-group. We preferred to state the proposition in a more restricted setting to remain in the geist of the rest of our paper.