Groups of p-rank 2 containing an isolated element of order p

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.

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.

The notation used is given in [10] or [3].

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

$$\begin{aligned} \mathcal {OUT}_G(R):=(N_G(R)/C_G(R)) /O_p(N_G(R)/C_G(R)). \end{aligned}$$

[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

$$\begin{aligned} {D=[D,b]\le [P,b]\cap P\le [P,a]P'O_{3'}(C_G(V))\cap P=[P,a]P'(O_{3'}(C_G(V))\cap P)=[P,a]P'}. \end{aligned}$$

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

$$\begin{aligned} (s_{2k}^{d^2\beta +\delta +d\gamma })^{ay}= & {} ((ss^d_1)^3)^{ay}=((ss^d_1)^{ay})^3=(((ss^d_1)^{-1}z)^{y})^3=((ss^d_1)^{-1})^3\\= & {} ((ss^d_1)^3)^{-1}=(s_{2k}^{d^2\beta +\delta +d\gamma })^{-1}. \end{aligned}$$

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. $\square$