Abstract
Let p be an odd prime and G be a finite group with \(O_{p'}(G)=1\) of p-rank at most 2 that contains an isolated element of order p. If \(x\notin Z(G)\), we show that \(F^*(G)\) is simple and we describe the structure of a Sylow p-subgroup P of \(F^*(G)\) as well as the fusion system \(\mathcal F_P(F^*(G))\) without using the classification of finite simple groups.
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1 Introduction
An element x of a finite group G is called isolated if and only if \(x^G\cap C_G(x)=\{x\}\). In 1966, George Glauberman [5] showed that \(G=O(G)\cdot C_G(x)\) if x is an isolated involution. A generalisation of his result for elements of odd prime order is a consequence of the classification of the finite simple groups. But there is still no proof revealing conceptual reasons or a deeper understanding of finite groups in terms of the subgroup structure. Even for the prime 3 the influence of isolated elements on finite groups is not understood. Together with Rebecca Waldecker, the author studied in [11] isolated elements of order 3. Under specific assumptions, they have been able to prove a version of the so-called \(Z_3^*\)-Theorem. One of these assumptions was the existence of an elementary abelian subgroup of order 27.
In this paper, we study a finite group G with \(O_{p'}(G)=1\) for an odd prime p such that G has an isolated element x of order p but does not contain any elementary abelian subgroup of order \(p^3\). We will first show that either \(x\in Z(G)\) or that \(F^*(G)\) is nonabelian simple and investigate the case where a Sylow p-subgroup of \(F^*(G)\) is abelian.
Then we assume that a Sylow p-subgroup P of \(F^*(G)\) is not abelian. The restriction on the p-rank of G implies that every subgroup of order \(p^3\) is generated by two elements. This allows to apply a theorem of Norman Blackburn [2] to obtain a list of possible isomorphism types for P. Most of the cases can be eliminated by more arguments regarding p-fusion and p-transfer in G. Assuming that P is not extraspecial of order \(p^3\) and exponent p, we deduce that P is a 3-group of maximal class and determine the possible isomorphism types of P with help of Chris Parker’s and Jason Semeraro’s results in [9]. More precisely we will prove the following theorem:
Theorem
Let G be a finite group and p be an odd prime such that \(O_{p'}(G)=1\). Suppose that G does not have an elementary abelian subgroup of order \(p^3\) and that there is some isolated element \(x\in G\) of order p such that \(x\notin Z(G)\).
Then \(F^*(G)\) is nonabelian simple and for every Sylow p-subgroup P of \(F^*(G)\), either \(x^G\cap P= \varnothing \) and P is cyclic or \(x^G\cap P\ne \varnothing \) and one of the following holds:
-
(a)
P is extraspecial of order \(p^3\) and exponent p.
-
(b)
\(p=3\), \(P=\langle s,s_1,...,s_{2k}\mid \forall i\in \{2,...,2k\}: s_i= [s_{i-1}, s],[s_1, s_{i}] = 1, \forall i\in \{2,...,2k-1\}: s^3_{i-1}s_i^3=s^{-1}_{i+1}\text { and }s^3=s_{2k-1}^3=s_{2k}^3=1\rangle \) is of maximal class and of order \(3^{2k+1}\) for some integer \(k\ge 2\).
We further use the classification of fusion systems on finite 3-groups of maximal class and rank 2 of [4] and [9] to see that P is normal in the fusion system induced by \(F^*(G)\) on P or that case (b) of the above theorem holds and the fusion system induced by \(F^*(G)\) on P is isomorphic to the fusion system of \(\mathrm {SL}^\varepsilon _3(q)\) for some \(\varepsilon \in \{1,-1\}\) and some prime power q such that \(3^{k} \mid q-\varepsilon \) but \(3^{k+1} \not \mid q-\varepsilon \) on a Sylow 3-subgroup.
2 Preliminaries
Definition 2.1
Let G be a finite group, p be a prime, and \(n\in {\mathbb {N}}\).
-
(a)
An element \(x\in G\) is called isolated in G if and only if \(x^G\cap C_G(x)=\{x\}\).
-
(b)
If G has an elementary abelian subgroup of order \(p^n\) but does not contain any of order \(p^{n+1}\), then G has p-rank n. We write \(r_p(G)=n\). We further say that G has rank n and write \(r(G)=n\) if G is a p-group.
-
(c)
A proper subgroup U of G is strongly p-embedded in G if and only if \(p\mid |U|\) but \(p\not \mid |U\cap U^g|\) for all \(g\in G\setminus U\).
Lemma 2.2
Let G be a finite group and p be a prime. Suppose that \(x\in G\) is an isolated element of order p in G. Then x is isolated in H for every subgroup H of G such that \(x\in H\). If further \(x\in P\in \mathrm {Syl}_p(G)\), then \(x\in Z(P)\) and if P is cyclic, then G has a normal p-complement.
Proof
The assertions follow from [11, Lemma 3.2 and 3.4]. \(\square \)
Lemma 2.3
Let P be a p-group for an odd prime p and \(\alpha \) be an automorphism of P of coprime order. Then the following hold:
-
(a)
\(P/P'=[P,\alpha ]P'/P'\times C_P(\alpha )P'/P'\).
-
(b)
If \(\alpha \) centralises \(\Omega _1(P)\), then \(\alpha \) centralises P.
Proof
Part (a) follows from [10, 8.2.2 and 8.4.2] and (b) is [6, 5.3.10]. \(\square \)
Lemma 2.4
Suppose that G is a finite group, p is an odd prime, P is a Sylow p-subgroup of G of rank 2 containing an isolated element x of order p. Then for all \(R\le P\) the group \((N_G(R)/C_G(R)) /O_p(N_G(R)/C_G(R))\) is cyclic of order dividing \(p-1\) or R has a subgroup that is extraspecial of order \(p^3\) and exponent p.
Proof
Let \(R\le P\) such that R does not have an extraspecial subgroup of order \(p^3\) and exponent p and set
[1, Corollary 14.4] provides a characteristic subgroup T of R of exponent p and class at most 2 such that every nontrivial \(p'\)-automorphism of R induces a nontrivial automorphism of T. Our assumption on R and \(r(T)\le r(P)=2\) imply that T does not have a subgroup of order \(p^3\) and so T is elementary abelian of order at most \(p^2\).
If T is cyclic, then [10, 2.2.5] yields that \(\mathcal {OUT}_G(T)\) is cyclic of order dividing \(p-1\). In the other case, if T is not cyclic, then from \(r(P)\le 2\) and \(x\in C_P(T)\), by Lemma 2.2, we see that \(x\in T\). Hence we deduce that \(N_G(T)/C_G(T)\) is isomorphic to a subgroup of \(C_p\rtimes C_{p-1}\), as x is isolated in G. So in both cases \(\mathcal {OUT}_G(T)\) is cyclic of order dividing \(p-1\). Since every nontrivial \(p'\)-automorphism of R induces a nontrivial automorphism of T, we conclude that also \(\mathcal {OUT}_G(R)\) is cyclic of order dividing \(p-1\). \(\square \)
Proposition 2.5
Let G be a finite group and p be an odd prime such that \(O_{p'}(G)=1\). Suppose that \(r_p(G)\le 2\) and that \(x\in G\) is isolated of order p such that \(x\notin Z(G)\).
Then \(F^*(G)\) is nonabelian simple.
Proof
First Lemma 2.2 yields that x centralises a Sylow p-subgroup P of G and so x centralises \(O_p(G)\).
Suppose for a contradiction that x centralises \(F^*(G)\). Then \(x\in Z(F^*(G))\) and so \(x^G\subseteq Z(F^*(G))\le C_G(x)\). From \(x^G\cap C_G(x)=\{x\}\), we see that \(x\in Z(G)\), a contradiction showing that x does not centralise \(F^*(G)\).
By \(O_{p'}(G)=1\), we obtain a component K of G such that \([K,x]\ne 1\) and \(p\mid |K|\). Let \(R:=P\cap K\in \mathrm {Syl}_p(K)\), then \(1\ne R\le P\le C_G(x)\) and Lemma 2.2 gives that \(R\langle x\rangle \) is not cyclic.
Suppose for a contradiction that \(Z(K)\ne 1\) and set \(Z_0:=\Omega _1(Z(K))\). Then \(Z(K)=O_p(G)\cap K\le R\) and \(x\notin Z(K)\le Z(F^*(G))\). From \(r(R\langle x\rangle )\le r_p(G)\le 2\) and \(x\in Z(R\langle x\rangle )\), we deduce that \(\Omega _1(R\langle x\rangle )=Z_0\times \langle x\rangle \) is elementary abelian of order \(p^2\). Moreover, Frobenius’s p-complement theorem [10, 7.2.4] provides a subgroup \(T\le R\) such that \(N_K(T)/C_K(T)\) is not a p-group. It follows that \(N_K(\Omega _1(T))/C_K(\Omega _1(T))\) is not a p-group by Lemma 2.3 (b). On the other hand, \(N_K(\Omega _1(T))\le K\le C_G(Z_0)\). We conclude that \(\Omega _1(T)\) is not equal to \(Z_0\) but contained in \(\Omega _1(R\langle x\rangle )=Z_0\times \langle x\rangle \). Let \(g\in N_K(\Omega _1(T))\setminus C_K(\Omega _1(T))\subseteq C_G(Z_0)\). Then there is some \(t\in \Omega _1(T)\le Z_0\times \langle x\rangle \) such that \(t^g\ne t\) and \(t\notin Z_0\). We may choose t such that \(t=yx\) for some \(y\in Z_0\). Then \(yx\ne (yx)^g=yx^g\) implies that \(x\ne x^g\). Since \(x^g=y^{-1}t\in \Omega _1(T)Z_0\le Z_0\times \langle x\rangle \le C_G(x)\), this is a contradiction showing that K is simple.
Suppose for a final contradiction that \(K\ne F^*(G)\). Then there is a subgroup \(E\le F^*(G)\) such that \(E\times K=F^*(G)\) and \(E\ne 1\) is subnormal in G. We set \(S:=E\cap P\) and observe \(S\ne 1\). Additionally \(SR=S\times R\) and from \(r_p(G)\le 2\), we deduce that S and R have rank 1 and so R and S are cyclic, as p is odd and by [10, 5.3.7]. Moreover, \(x\in C_G(R\times S)\) and so \(r_p(G)\le 2\) gives that \(x\in R\times S\). Hence there are elements \(a\in R\) and \(b\in S\) such that \(x=ab\). Since \(x\notin C_G(K)\ge S\), we see that \(a\ne 1\). As R is cyclic and a Sylow p-subgroup of the simple group K, Burnside’s p-complement theorem (see [10, 7.2.1]) provides an element \(g\in N_K(R)\le C_G(E)\) that has order coprime to p and does not centralise R. Lemma 2.3 (b) then gives that g does not centralise a. Thus \(x^g=a^gb\in RS\le C_G(x)\) but \(x^g\ne x\). This is a contradiction. \(\square \)
Definition 2.6
Let p be a prime, G a finite group, \(P\in \mathrm {Syl}_p(G)\), and \({\mathcal {F}}:={\mathcal {F}}_P(G)\) the fusion system of G on P. Let further Q be a subgroup of P.
-
(a)
Q is called fully \({\mathcal {F}}\)-normalised if and only if \(N_P(Q)\in \mathrm {Syl}_p(N_G(Q))\).
-
(b)
Q is called \({\mathcal {F}}\)-centric if and only if \(C_G(Q)=Z(Q)\times O_{p'}(C_G(Q))\).
-
(c)
Q is called \({\mathcal {F}}\)-essential if and only if \(N_G(Q)/QC_G(Q)\) contains a strongly p-embedded subgroup and Q is \({\mathcal {F}}\)-centric.
-
(d)
Q is called \({\mathcal {F}}\)-radical if and only if \(O_p(N_G(Q)/QC_G(Q))=1\).
-
(e)
Q is called \({\mathcal {F}}\)-Alperin if and only if Q is fully \({\mathcal {F}}\)-normalised, \({\mathcal {F}}\)-centric, and \({\mathcal {F}}\)-radical.
Lemma 2.7
Let G be a finite group, p be a prime, and P be a Sylow p-subgroup of G. If \([D,N_G(D)]\le [P,N_G(P)]\) for all \({\mathcal {F}}_P(G)\)-essential subgroups D of G, then we have \(P\cap G'=[P,N_G(P)]\).
Proof
By Alperin’s fusion theorem [3, 4.51] and the focal subgroup theorem [10, 7.1.3], the group \(P\cap G'\) is generated by \([P, N_G(P)]\) and \([D,N_G(D)]\) where D runs through all \({\mathcal {F}}_P(G)\)-essential subgroups. Thus \(P\cap G'= [P, N_G(P)]\) by the assumptions. \(\square \)
3 The structure of a Sylow p-subgroup
Proposition 3.1
Let G be a finite group and p be an odd prime such that \(O_{p'}(G)=1\). Suppose that \(r_p(G)\le 2\) and that \(x\in G\) is isolated of order p such that \(x\notin Z(G)\). Then \(x\notin F^*(G)\) if and only if \(F^*(G)\) has abelian Sylow p-subgroups if and only if \(F^*(G)\) has cyclic Sylow p-subgroups.
Proof
We set \(K:=F^*(G)\). Then K is nonabelian simple by Proposition 2.5. Let further S be a Sylow p-subgroup of \(K\langle x\rangle \) containing x. Then \(P:=S\cap K\) is a Sylow p-subgroup of K and \(P\langle x\rangle =S\).
Suppose first that \(x\notin K\). Then \(S=P\times \langle x\rangle \), as x centralises every Sylow p-subgroup of G that contains x by Lemma 2.2. Since \(r_p(G)\le 2\), we see that P does not contain an elementary abelian subgroup of order \(p^2\) and so P is cyclic. If P is cyclic, then P is abelian.
Let finally P be abelian. Then again Lemma 2.2 shows that S is abelian. We choose an element \(u\in S\) of maximal order. Then [10, Lemma 2.1.2] provides a subgroup R of S such that \(S=\langle u\rangle \times R\). As \(r(P\langle x\rangle )\le 2\), we see that R is cyclic. In particular, \(S/\langle u\rangle \) and S/R are cyclic groups. Since \(\langle u\rangle \cap R=1\), there is some \(Q\in \{\langle u\rangle ,R\}\) such that \(x\notin Q\). Then x is an element of \(S\setminus Q\) of order p. In addition every G-conjugate of x in S is equal to x since x is isolated in G. We apply [7, Lemma 15.18] to conclude that \(x\notin O^p(K\langle x\rangle )=K\). \(\square \)
Lemma 3.2
Let P be a finite p-group for an odd prime p such that \(r(P)\le 2\).
Then one of the following holds:
-
(a)
P is a 3-group of maximal class,
-
(b)
\(\Omega _1(P)\) is extraspecial of order \(p^3\) and exponent p and \(P/\Omega _1(P)\) is cyclic,
-
(c)
P is metacyclic, or
-
(d)
\(|P|\le p^4\).
Proof
We may suppose that \(|P|\ge p^5\) and let R be a normal subgroup of P of order \(p^3\). Then R is not elementary abelian and so \(\Phi (R)\ne 1\). Thus \(|R/\Phi (R)|< |R|=p^3\) and we deduce from [10, 5.2.5] that R is generated by two elements.
Altogether we may apply [2, Theorem 4.1]. \(\square \)
Hypothesis 3.3
Let G be a finite nonabelian simple group and p be an odd prime. Suppose that \(r_p(G)\le 2\) and that \(x\in G\) of order p is isolated in G. Let further \(P\in \mathrm {Syl}_p(G)\) be such that \(x\in P\) and assume that P is not extraspecial of order \(p^3\) and exponent p.
Lemma 3.4
Assume Hypothesis 3.3 and let Q be an extraspecial normal subgroup of P of order \(p^3\) and exponent p. Then P/Q is not cyclic.
Proof
From Hypothesis 3.3, we see that \(P\ne Q\) and Lemma 2.2 as well as \(r(P)\le 2\) imply that \(\langle x\rangle =Z(Q)\). Moreover, [1, Lemma 1.4] provides a normal subgroup V of P that is contained in Q and is elementary abelian of order \(p^2\). Then \(Q\nleq C_P(V)\) and from \(|\mathrm {Aut}(V)|_p=p\), we see that \(C_P(V)\) is a maximal subgroup of \(P=QC_P(V)\).
We suppose for a contradiction that P/Q is cyclic and choose \(w\in C_P(V)\setminus V\) such that \(w^p\in Q\). From \(r(P)\le 2\) and \(w\in C_P(V)\setminus V\), we get that w does not have order p. We conclude that \(1\ne w^p\in C_P(V)\cap Q=V\). Moreover, \(\langle w\rangle V\) is a maximal and hence normal subgroup of \(\langle w\rangle Q\). So \(\Phi (\langle w\rangle V)=\langle w^p\rangle \) is \(\langle w\rangle Q\)-invariant. Now \(w^p\in Q\) implies that \(\langle w^p\rangle \) is normal in Q and so \(\langle w^p\rangle =Z(Q)=\langle x\rangle \).
We may assume that \(w^p=x\). Since x is isolated in G, no G-conjugate of a power of \(w^p\) lies in \(P\setminus Q\). Further \(w^g\notin Q\) for all \(g\in G\), as w does not have order p and Q has exponent p. From \(w\in G=O^p(G)\) and [7, Proposition 15.15], we obtain an element \(g\in G\) such that \(w^g=w^{k}z\) for some \(z\in Q\) and \(k\in {\mathbb {N}}\) such that p does divide neither k nor \(k-1\) and such that \(C_P(w^g)\in \mathrm {Syl}_p(C_G(w^g))\).
If we had \([w^k,z]=1\), then \(x^k=(w^p)^k=w^{kp}z^p=(w^kz)^p=x^g\in C_G(x)\cap x^G=\{x\}\) would be a contradiction. Hence \(w\in C_P(V)\setminus C_P(z)\) and so \(Q=\langle z,V\rangle \). We further deduce that \(C_P(z)=C_Q(z)\) from \(\langle w\rangle Q/Q=\Omega _1(P/Q)\) and so \(C_P(Q)= C_P(z)\cap C_P(V)\le C_Q(z)\cap C_P(V)=Z(Q)\). Now [12] gives that \(|N_G(Q)/QC_G(Q)|\) divides \(p(p^2-1)\), as \(Z(Q)=\langle x\rangle \) is centralised by \(N_G(Q)\). Consequently \(P/Q=N_P(Q)/QC_P(Q)\) has order p and so \(P=Q\langle w\rangle \) has order \(p^4\). By [12], the element w does not normalise \(\langle x,z\rangle \). Altogether we have \(\langle w^g,w\rangle =\langle z,w\rangle =Q\langle w\rangle =P\), implying that \(C_P(w)\cap C_P(w^g)\le Z(P)=\langle x\rangle \). This shows that \(|P:C_P(w^g)|\ge |C_P(w):C_P(w)\cap C_P(w^g)|=|V\langle w\rangle :\langle x\rangle |=p^2\), entailing that \(C_P(w^g)\) has order \(p^2\). But \(C_P(w^g)\in \mathrm {Syl}_p(C_G(w^g))\) and so \(|C_P(w^g)|\ge |(C_P(w))^g|\ge |(V\langle w\rangle )^g|=p^3\). This is a contradiction. \(\square \)
Lemma 3.5
Assume Hypothesis 3.3. Then P has a unique elementary abelian normal subgroup of order \(p^2\).
Proof
As in the lemma before, [1, Lemma 1.4] provides a normal elementary abelian subgroup V of P of order \(p^2\). Let \(W\unlhd P\) be elementary abelian of order \(p^2\) and suppose for a contradiction that \(V\ne W\). From \(r(P)\le 2\) and \([V,W]\le V\cap W\), we see that \(V\gneq V\cap W\ne 1\) and so \(|VW|=\frac{|V|\cdot |W|}{|V\cap W|}=p^3\) by [10, 1.1.6]. We set \(Q:=VW\). Then \(Q=\Omega _1(VW)\) and so \(r(Q)\le r(P)=2\) implies that Q is not abelian. Thus Q is extraspecial of order \(p^3\) by [8, I.14.10] and of exponent p, as \(Q=\Omega _1(VW)\). Every \(y\in P\) normalises Q, V, and W and so y normalises every maximal subgroup of Q. In particular, y centralises the elementary abelian group \(Q/\Phi (Q)=Q/Z(Q)\) and hence \([P,Q]\le Z(Q)\). Thus [12] implies that \(P=Q\cdot C_P(Q)\).
For all \(z\in C_P(Q)\) of order p, we see that \(\langle V,z\rangle \) and \(\langle W,z\rangle \) are elementary abelian. Then \(r(P)=2\) yields that \(z\in V\cap W\le Z(Q)\). Thus \(C_P(Q)\) has a unique minimal subgroup and is consequently cyclic, as p is odd. This contradicts Lemma 3.4. \(\square \)
Lemma 3.6
Suppose that Hypothesis 3.3 holds and that P does not have an extraspecial subgroup of order \(p^3\) and exponent p. Then there is some element \(a\in N_G(P)\) with \(a^{p-1}\in O_{p'}(C_G(P))\) that induces a fixed-point-free automorphism on \(P/P'\); moreover \(C_P(a)\) is cyclic.
Proof
We investigate the fusion system \({\mathcal {F}}_P(G)\) and let D be an \({\mathcal {F}}_P(G)\)-radical subgroup. Then \(O_p(N_G(D)/C_G(D))=DC_G(D)/C_G(D)\), our assumption and Lemma 2.4 imply that \(|N_G(D)/DC_G(D)|\) divides \(p-1\).
If D is \({\mathcal {F}}_P(G)\)-essential subgroup of P, then D is also \({\mathcal {F}}_P(G)\)-radical (see [3, p. 119]) and \({\mathcal {F}}_P(G)\)-centric. It follows that \(N_P(D)\le DC_G(D)=DZ(D)O_{p'}(C_G(D))=D O_{p'}(C_G(D))\) and hence \(D=P\). This is a contradiction. In particular, there does not exists any \({\mathcal {F}}_P(G)\)-essential subgroup of P. From Lemma 2.7, we deduce that \(P=P\cap G'=[P,N_G(P)]\) and so \(H:=N_G(P)\ne C_G(P)P\).
As P is \({\mathcal {F}}_P(G)\)-radical, the above argument provides some \(a\in H\setminus O_{p'}(H)P\) of order prime to p such that \(H=O_{p'}(H)P\langle a\rangle \) and \(a^{p-1}\in O^p(PC_G(P))=O_{p'}(H)\). We conclude that \(P=[P,H]=P'\cdot [P,a]\) and hence that a acts fixed-point-freely on the abelian group \(P/P'\) by Lemma 2.3 (a).
Let now V be the unique normal elementary abelian subgroup of order \(p^2\) of P which exists by Lemma 3.5. Then \(\Omega _1(C_P(V))=V\), as \(r(P)\le 2\), and \(C_P(V)\) is a maximal subgroup of P. We conclude that \(C_P(a)\le P'\lneq C_P(V)\) and so \(\Omega _1(C_P(a))\le \Omega _1(C_P(V))=V\). If \(C_P(a)\) was not cyclic, then a would centralise \(\Omega _1(C_P(V))=V\) and from Lemma 2.3 (b), we would deduce that \(C_P(V)\le C_P(a)\), a contradiction. \(\square \)
Lemma 3.7
Suppose that Hypothesis 3.3 holds. Then P is not metacyclic.
Proof
Suppose for a contradiction that P is metacyclic. Then P does not have an extraspecial subgroup of order \(p^3\) and exponent p and so Lemma 3.6 provides some \(a\in N_G(P)\) with \(a^{p-1}\in C_G(P)\) that acts fixed-point-freely on \(P/P'\) and hence on \(\Omega _1(P/P')\) of order \(p^2\). Thus [8, Satz II 3.10] and \(o(a)\mid p-1\) provide at least one a-invariant subgroup of order p of \(\Omega _1(P/P')\). In addition Maschke’s theorem (see for example [10, 8.4.6]) gives a second one. The full pre-images of these a-invariant subgroups are maximal subgroups of the full pre-image U of \(\Omega _1(P/P')\). As P is metacyclic, U has a maximal cyclic subgroup. Hence by [1, Theorem 1.2 (a)], all but at most one maximal subgroups of U are cyclic, as p is odd. Altogether U has a maximal subgroup R, that is a-invariant, cyclic, and contains \(P'\).
Since a acts fixed-point-freely on \(P/P'\), a does not centralise \(R/P'\) and hence not R. From Lemma 2.3 (b), we see that a does not centralise \(\Omega _1(R)=\Omega _1(P')\) and so \(C_P(a)\le C_{P'}(a)=1\). This contradicts \(x\in C_P(a)\), as x is isolated in G. \(\square \)
Lemma 3.8
Suppose that Hypothesis 3.3 holds. Then \(|P|\ge p^5\).
Proof
From Lemma 3.7 and Hypothesis 3.3, it firstly follows that \(|P|\ge p^4\).
Suppose for a contradiction that \(|P|= p^4\). If P contained an extraspecial subgroup Q of order \(p^3\) and exponent p, then Q would be a maximal subgroup of P and hence normal in P with cyclic factor group, contradicting Lemma 3.4. We conclude that every proper subgroup R of P is metacyclic. It follows from [2, Theorem 3.2] that P is a 3-group of class 3 and order \(3^4\), as P itself is not metacyclic by Lemma 3.7. In particular, we have \(p=3\). If V is a normal elementary abelian subgroup of P of order 9, that exists by Lemma 3.5, then P/V has order 9 and so \(P'\le V\). It follows that \(P'\) is elementary abelian of order 9, as P has class 3.
In addition Lemma 3.6 provides an element \(a\in N_G(P)\) that induces a fixed-point-free automorphism of order 2 on \(P/P'\) and such that \(C_P(a)\) is cyclic. Since x is isolated in P, we see that \(x\in C_P(a)\le P'=V\) and hence \(C_P(a)=\langle x\rangle =Z(P)\unlhd P\). Altogether a induces a fixed-point-free automorphism of order 2 on the nonabelian group \(P/Z(P)=P/C_P(a)\). This contradicts [10, 8.1.10]. \(\square \)
Lemma 3.9
Suppose that Hypothesis 3.3 holds and that P is a 3-group of maximal class. Then \(N_G(P)=C_G(P)P\langle a\rangle \) where \(a^2\in C_G(P)\) and a acts fixed-point-freely on \(P/P'\) and \(|P|=3^{2k+1}\) for some integer \(k\ge 2\).
Proof
Let \(|P|=3^n\). Then Lemma 3.8 implies that that \(n\ge 5\). As P has maximal class, we see that \(\langle x\rangle =Z(P)\) by Lemma 2.2. Now, [9, Proposition 3.3] describes \(\mathrm {Aut}(P)\). Let \(a\in N_G(P)\), then the proposition provides \(e,f\in \{1,2\}\) such that \(x^a=x^{e^{n-2}f}\) by [9, Lemma 3.4]. As x is isolated in G, we see \(1\equiv e^{n-2}\cdot f\) mod 3. So if n is even, then \(e^{n-2}\equiv 1\) mod 3 and so \(f=1\). If n is odd, then \(e^{n-2}\equiv e\) mod 3 and so \(f=e\). In both cases, we conclude again from [9, Proposition 3.3] that \(|N_G(P)/C_G(P)|_2\le 2\) and hence that \(N_G(P)/PC_G(P)\) is cyclic of order at most 2. Let \(a\in N_G(P)\) be a possibly trivial 2-element such that \(N_G(P)=C_G(P)P\langle a\rangle \) and \(a^2\in C_G(P)\). We want to apply Lemma 2.7. Therefore we first observe that \([P,N_G(P)]=[P, C_G(P)P\langle a\rangle ]=P'[P,a]\).
Suppose now that D is an \({\mathcal {F}}_P(G)\)-essential subgroup. Then \(N_G(D)/DC_G(D)\) contains a strongly 3-embedded subgroup and so D has a subgroup that is extraspecial of order 27 and exponent 3 by Lemma 2.4. Thus [9, Lemma 4.1 and Lemma 4.2] yield that D itself is extraspecial of order \(3^3\) and exponent 3 and determine \(N_G(D)/DC_G(D)\). From \(r(P)\le 2\) and \(x\in Z(P)\), by Lemma 2.2, we have \(x\in Z(D)\) and so \(N_G(D)/DC_G(D)\) is isomorphic to \(\mathrm {Sp}_2(3)=\mathrm {SL}_2(3)\), as x is isolated in G. By Lemma 3.5, the group P has a unique normal elementary abelian subgroup V of order 9. Then \(C_P(V)\) is a maximal subgroup of P and from \(V=\Omega _1(C_P(V))\), we deduce that \(V\le D\).
Let \(b\in N_G(D)\) be such that b induces an automorphism of order 2 on D. Then b acts on D such that it normalises but does not centralise any elementary abelian subgroup of order 9. It follows that \(b\in N_G(V)\setminus C_G(V)\) and so that \(N_G(V)/C_G(V)\cong S_3\). By a Frattini argument, we have \(N_G(V)=C_G(V)P\cdot N_G(P)\), so we get that \(C_G(P)P\langle a\rangle =N_G(P)\nleq C_G(V)P\). In particular, a does not centralise but normalise the characteristic subgroup V of P. Since \(|N_G(V)/C_G(V)P|=2\), there is some \(c\in C_G(V)P\) such that \(b=ca\).
From the Frobenius normal p-complement theorem [10, 7.2.4], we get \(V=\Omega _1(C_P(V))\), and Lemma 2.3 (b), we see that \(C_G(V)\) has a normal 3-complement. So \(C_G(V)P\le O_{3'}(C_G(V))P\) and it follows that \([P,c]\le [P,PO_{3'}(C_G(V))]\le P'O_{3'}(C_G(V))\). As \(a\in N_G(P)\le N_G(V)\), we further see that a normalises \(P'O_{3'}(C_G(V))\) and hence \([P,b]=[P,ca]=[P,a][P,c]^a\le [P,a]P'O_{3'}(C_G(V))\). By [10, 1.1.11], we get
We summarise that \([D,N_G(D)]\le [P,a]P'\le [P,N_G(P)]\) for every \({\mathcal {F}}\)-essential subgroup D of P. Hence Lemma 2.7 shows that \(P=G'\cap P=[P,N_G(P)] =P'[P, a]\). Since a acts coprimely, we infer that a acts fixed-point-freely on \(P/P'\).
Altogether a inverts \(P/P'\) and the parameters e and f in [9, Proposition 3.2] for a are both equal to 2. Finally our argument at the beginning of this proof shows that n is odd. \(\square \)
Lemma 3.10
If Hypothesis 3.3 holds and P is a 3-group of maximal class, then \(P=\langle s,s_1,...,s_{2k}\mid \forall i\in \{2,...,2k\}: s_i= [s_{i-1}, s],[s_1, s_{i}] = 1, \forall i\in \{2,...,2k-1\}: s^3_{i-1}s_i^3=s^{-1}_{i+1}\text { and }s^3=s_{2k-1}^3=s_{2k}^3=1\rangle \) has order \(3^{2k+1}\) for some \(k\ge 2\).
Proof
Let \(|P|=3^n\). Then Lemma 3.9 implies that \(n=2k+1\) for some integer \(k\ge 2\). From [9], we obtain some \(\beta ,\gamma ,\delta \in \{0,1,2\}\) such that \(P:=\langle s,s_1,...,s_{2k}\mid R1, R2, R3, R4, R5, R6\rangle \), where the relations are as follows:
R1: \(\forall i\in \{2,...,2k\}: s_i=[s_{i-1},s]\);
R2: \(\forall i\in \{3,...,2k\}: [s_1, s_i] = 1\);
R3: \(\forall i\in \{2,...,2k\}: s_i^3s_{i+1}^3s_{i+2}=1\), where \(s_{2k+1}=s_{2k+2}=1\) by definition;
R4: \([s_1,s_2]=s_{2k}^\beta \); R5: \(s^3_1s_2^3s_3=s_{2k}^\gamma \); and R6: \(s^3=s_{2k}^\delta \).
To prove the assertion, we need to verify that \(\beta =\gamma =\delta =0\).
With regard to Lemma 3.9, let \(a\in N_G(P)\) be such that \(a^{2}\in C_G(P)\) and such that a induces a fixed-point-free automorphism on \(P/P'\). In particular, a inverts \(P/P'\). Then [9, Lemma 2.1] states that \(P'=[\langle s_1,...,s_{2k}\rangle ,P]\) and \(Z(P)=\langle s_{2k}\rangle =\langle x\rangle \).
For all \(d\in \{0,1,2\}\), we see that \(ss^d_1\notin P'\) and so there is some \(z\in P'\) such that \((ss^d_1)^a=(ss^d_1)^{-1}z\). As \((ss_1^d)^{-1}\notin \langle s_1,...,s_{2k}\rangle \), [9, Lemma 2.4] provides some \(y\in P\) such that \(((ss^d_1)^{-1}z)^y=(ss^d_1)^{-1}\). A calculation before [9, equation (3.3)] gives that \((ss^d_1)^3=s_{2k}^{d^2\beta +\delta +d\gamma }\). Altogether we obtain that
But \(s_{2k}\in \langle x\rangle \) is isolated in G and so \(d^2\beta +\delta +d\gamma \) is divisible by 3. This is true for all \(d\in \{0,1,2\}\) and so \(\delta \) is divisible by 3, which gives \(\delta =0\). In addition we get that \(\beta +\gamma \) and \(\beta -\gamma \) are divisible by 3 and so is their sum \(2\beta \) and their difference \(2\gamma \). From \(\beta ,\gamma \in \{0,1,2\}\), we obtain that \(\beta =\gamma =0\). \(\square \)
Proof of the main Theorem
We investigate \(K:=F^*(G)\) and let \(P\in \mathrm {Syl}_p(K)\). Then K is nonabelian simple by Proposition 2.5 and does not contain any elementary abelian subgroup of order \(p^3\), as G does not. In particular, P satisfies the hypothesis of Lemma 3.2.
If \(x\notin K\), then \(x^G\cap P=\varnothing \) and Proposition 3.1 implies that P is cyclic. So we may suppose that \(x\in K\). Then x is isolated in K by Lemma 2.2.
If K does not have extraspecial Sylow p-subgroups of order \(p^3\) and exponent p, then Hypothesis 3.3 holds and so Lemma 3.2 yields together with Lemma 3.4, Lemma 3.7, and Lemma 3.8 that P is a 3-group of maximal class. Finally Lemma 3.10 provides the assertion. \(\square \)
4 The fusion system induced by G
Theorem 4.1
Let G be a finite group and p be an odd prime such that \(O_{p'}(G)=1\). Suppose that G does not have an elementary abelian subgroup of order \(p^3\) and that there is some isolated element \(x\in G\) of order p such that \(x\notin Z(G)\).
Then every Sylow p-subgroup P of \(F^*(G)\) is normal in \({\mathcal {F}}_P(F^*(G))\) or part (b) of the main Theorem is true and \({\mathcal {F}}_P(F^*(G))\) is isomorphic to \({\mathcal {F}}_P(H)\), where \(H\cong \mathrm {SL}^\varepsilon _3(q)\) for some \(\varepsilon \in \{1,-1\}\) and some prime power q such that \(3^{k} \mid q-\varepsilon \) but \(3^{k+1} \not \mid q-\varepsilon \).
Proof
Let \(P\in \mathrm {Syl}_p(F^*(G))\) and set \({\mathcal {F}}:={\mathcal {F}}_P(F^*(G))\). If P is cyclic, then it follows from a result of Burnside (see [10, 7.1.5]) that P is normal in \({\mathcal {F}}\).
From now on, we may suppose that \(x\in P\). In particular, Proposition 3.1 gives that P is not abelian. Further [3, Theorem 5.39 (v)] implies that P is normal in \({\mathcal {F}}\) if and only if P is the only \({\mathcal {F}}\)-Alperin subgroup. Let \(R\le P\) be abelian, then \(P\ne R\). Using Lemma 2.4, we see that \((N_G(R)/C_G(R)) /O_p(N_G(R)/C_G(R))\) is cyclic of order dividing \(p-1\). If R is \({\mathcal {F}}\)-radical and \({\mathcal {F}}\)-centric, then we deduce from \(O_p(N_G(R)/C_G(R)) = RC_G(R)/C_G(R) = 1\) that \(N_P(R)\le C_P(R)\le R\), so that \(R = P\), which is a contradiction. This shows that R is not \({\mathcal {F}}\)-Alperin.
If case (a) of the main Theorem holds, then P is extraspecial of order \(p^3\) and exponent p. So every proper subgroup of P is abelian. We deduce that P is the only \({\mathcal {F}}\)-Alperin subgroup, whence P is normal in \({\mathcal {F}}\).
Consequently, if P is not normal in \({\mathcal {F}}\), then we see that part (b) of the main theorem holds. In addition P has a proper \({\mathcal {F}}\)-Alperin subgroup R. We check the possibilities of \({\mathcal {F}}\) in [4, Table 6 of Theorem 5.10] which is correct in our case by [9, Theorem 1]. We already saw that abelian subgroups of P are not \({\mathcal {F}}\)-Alperin and so the groups denoted by \(V_0\), \(V_1\), \(V_{-1}\), and \(\gamma _1\) are not \({\mathcal {F}}\)-Alperin. Finally if \(R\le P\) is extraspecial of order \(p^3\) and exponent p, then \(x\in Z(R)\), as R has rank 2. So we see from [12] that \( \mathrm {Aut}_{{\mathcal {F}}}(R)\) is a subgroup of \(\mathrm {SL}_2(3)\). Since \({\mathcal {F}}\) is not exotic, we see that only the third row may occur and so our theorem is true. \(\square \)
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Toborg, I. Groups of p-rank 2 containing an isolated element of order p. Arch. Math. 118, 349–359 (2022). https://doi.org/10.1007/s00013-022-01703-7
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DOI: https://doi.org/10.1007/s00013-022-01703-7