Exponent of a finite group of odd order with an involutory automorphism

TL;DR

This paper establishes bounds on the exponent of the subgroup generated by elements of the form [g,φ] in a finite odd-order group with an involutory automorphism, linking these bounds to properties of the fixed point subgroup.

Contribution

It provides new bounds on the exponent of [G,φ] based on the structure and properties of the fixed point subgroup G_φ and elements in G_{-φ}, extending understanding of automorphism actions.

Findings

Bound on exponent when G_φ is nilpotent of class c

Bound on exponent when G_φ has rank r

Results depend on properties of elements in G_{-φ}

Abstract

Let $G$ be a finite group of odd order admitting an involutory automorphism $\phi$. We obtain two results bounding the exponent of $[G,\phi]$. Denote by $G_{-\phi}$ the set $\{[g,\phi]\,\vert\, g\in G\}$ and by $G_{\phi}$ the centralizer of $\phi$, that is, the subgroup of fixed points of $\phi$. The obtained results are as follows.1. Assume that the subgroup $\langle x,y\rangle$ has derived length at most $d$ and $x^e=1$ for every $x,y\in G_{-\phi}$. Suppose that $G_\phi$ is nilpotent of class $c$. Then the exponent of $[G,\phi]$ is $(c,d,e)$-bounded.2. Assume that $G_\phi$ has rank $r$ and $x^e=1$ for each $x\in G_{-\phi}$. Then the exponent of $[G,\phi]$ is $(e,r)$-bounded.

Full text

1. Assume that the subgroup $\langle x,y\rangle$ has derived length at most $d$ and $x^{e}=1$ for every $x,y\in G_{-\phi}$. Suppose that $G_{\phi}$ is nilpotent of class $c$. Then the exponent of $[G,\phi]$ is $(c,d,e)$-bounded.

2. Assume that $G_{\phi}$ has rank $r$ and $x^{e}=1$ for each $x\in G_{-\phi}$. Then the exponent of $[G,\phi]$ is $(e,r)$-bounded.

1. Introduction

Let $G$ be a finite group of odd order admitting an involutory automorphism $\phi$. Here the term “involutory automorphism” means an automorphism $\phi$ such that $\phi^{2}=1$. We denote by $G_{-\phi}$ the set $\{[g,\phi]\,|\,g\in G\}$ and by $G_{\phi}$ the centralizer of $\phi$, that is, the subgroup of fixed points of $\phi$. The subgroup generated by $G_{-\phi}$ will be denoted by $[G,\phi]$. It is well-known that $[G,\phi]$ is normal in $G$ and $\phi$ induces the trivial automorphism of $G/[G,\phi]$. The following theorem was proved in [1]:

Let $p$ be an odd prime and $G$ a finite $p$-group admitting an involutory automorphism $\phi$ such that $G_{\phi}$ is nilpotent of class $c$ and $x^{p}=1$ for each $x\in G_{-\phi}$. Suppose that the derived length of $G$ is at most $d$. Then the nilpotency class of $[G,\phi]$ is $(c,d,p)$-bounded.

Throughout this note we use the term “$(a,b,c\dots)$-bounded” to mean “bounded from above by some function depending only on the parameters $a,b,c\dots$”. An immediate corollary of the above theorem is that under its hypotheses the exponent of $[G,\phi]$ is $(c,d,p)$-bounded as well. Recall that the exponent of a group $K$ is the minimal number $e$ such that $x^{e}=1$ for each $x\in K$. In this note we address the question whether one can relax the hypotheses of the theorem while still being able to conclude that the exponent of $[G,\phi]$ is bounded in terms of relevant parameters. Let $\langle X\rangle$ denote the subgroup generated by the set $X$. We obtain the following result.

Theorem 1.1**.**

Let $c,d,e$ be nonnegative integers and $G$ a finite group of odd order admitting an involutory automorphism $\phi$ such that $G_{\phi}$ is nilpotent of class $c$ and $x^{e}=1$ for each $x\in G_{-\phi}$. Suppose that the subgroup $\langle x,y\rangle$ has derived length at most $d$ for every $x,y\in G_{-\phi}$. Then the exponent of $[G,\phi]$ is $(c,d,e)$-bounded.

We do not know whether or not all hypotheses in this theorem are necessary. It is conceivable that the exponent of $[G,\phi]$ in Theorem 1.1 can be bounded only in terms of $c$ and $e$, or only in terms of $d$ and $e$. This seems to be a difficult problem.

Recall that a finite group $G$ is said to be of rank $r$, if $r$ is the least number such that any subgroup of $G$ is $r$-generator. Our next theorem shows that the exponent of $[G,\phi]$ can be bounded in terms of $e$ and the rank of $G_{\phi}$. In this case we do not even need to assume that $G_{\phi}$ is nilpotent.

Theorem 1.2**.**

Let $e$ and $r$ be nonnegative integers and $G$ a finite group of odd order admitting an involutory automorphism $\phi$ such that $G_{\phi}$ has rank $r$ and $x^{e}=1$ for each $x\in G_{-\phi}$. Then the exponent of $[G,\phi]$ is $(e,r)$-bounded.

We do not know if our results can somehow be generalized to the cases where $\phi$ is not necessarily of order two, or the assumption that $(|G|,|\phi|)=1$ is dropped from the hypotheses.

Throughout the paper the Feit–Thompson Theorem [2] is used without explicit references.

2. Proof of Theorem 1.1

The next lemma is a collection of well-known facts about involutory automorphisms. In the sequel we will frequently use it without any reference.

Lemma 2.1**.**

If $G$ is a finite group of odd order admitting an involutory automorphism $\phi$, then

1. (1)

$G=G_{\phi}G_{-\phi}=G_{-\phi}G_{\phi}$, and each element $x\in G$ can be written uniquely in the form $x=gh$, where $g\in G_{-\phi}$ and $h\in G_{\phi}$; 2. (2)

If $N$ is any $\phi$-invariant normal subgroup of $G$ we have $(G/N)_{\phi}=G_{\phi}N/N$, and $(G/N)_{-\phi}=\{gN;g\in G_{-\phi}\}$; 3. (3)

If $N$ is a $\phi$-invariant normal subgroup of $G$ such that either $N=N_{-\phi}$ or $N=N_{\phi}$, then $[G,\phi]$ centralizes $N$; 4. (4)

The normal closure of $G_{\phi}$ contains $G^{\prime}$; 5. (5)

$G_{\phi}$* normalizes the set $G_{-\phi}$.*

Proposition 2.2**.**

Let $G$ be a finite group of odd order admitting an involutory automorphism $\phi$ and assume that $G$ is soluble with derived length $d$. Suppose that $G=[G,\phi]$ and $(G_{-\phi})^{e}=1$. Then $G$ has $(d,e)$-bounded exponent.

Proof.

We use induction on $d$. If $d=1$, the result is obvious, so we assume that $d\geq 2$. Let $M=G^{(d-1)}$ be the last nontrivial term of the derived series of $G$. By induction, the exponent of $G/M$ is $(d,e)$-bounded. Since $M=M_{-\phi}M_{\phi}$ and $(M_{-\phi})^{e}=1$, taking into account that $M$ is abelian, we deduce that $M^{e}\leq M_{\phi}$. In view of Lemma 2.1.3, we conclude that $M^{e}\leq Z(G)$. Hence, the exponent of $G/Z(G)$ is $(d,e)$-bounded. A theorem of Mann [8] now guarantees that $G^{\prime}$ has $(d,e)$-bounded exponent. Since $G$ is generated by elements of order dividing $e$, the result follows. ∎

Lemma 2.3**.**

Let $G$ be a finite group of odd order with an involutory automorphism $\phi$ such that $G=[G,\phi]$. Let $S$ be the set of those elements $h\in G_{\phi}$ for which there exist $x,y\in G_{-\phi}$ such that $h\in\langle x,y\rangle$. Then $G_{\phi}=\langle S\rangle$.

Proof.

Set $H=\langle S\rangle$. Obviously, $H\leq G_{\phi}$. So we need to prove that $G_{\phi}\leq H$. Choose $h\in G_{\phi}$. Since $G=[G,\phi]$, we can write $h=g_{1}\cdots g_{n}$ with $g_{i}\in G_{-\phi}$. We wish to prove that $h\in H$. This will be shown by induction on $n$. If $n\leq 2$, then obviously $h\in H$, so assume that $n\geq 3$. Let $K=\langle g_{n-1},g_{n}\rangle$ and note that $K$ is $\phi$-invariant. Write $g_{n-1}g_{n}=g_{0}h_{0}$ where $g_{0}\in K_{-\phi}$ and $h_{0}\in K_{\phi}$. Observe that $K_{\phi}\leq H$ and therefore $h_{0}\in H$. We have $h=g_{1}\cdots g_{n-2}g_{0}h_{0}$ and $hh_{0}^{-1}=g_{1}\cdots g_{n-2}g_{0}$. By induction we get $hh_{0}^{-1}\in H$, whence $h\in H$. Therefore, we conclude that $H=G_{\phi}$. ∎

The following theorem was proved in [11] (see also [12]). Its proof is based on a Lie-theoretical result of Zelmanov [13].

Theorem 2.4**.**

Let $e$ be a positive integer and $G$ a finite group of odd order admitting an involutory automorphism $\phi$ such that all elements in $G_{\phi}\cup G_{-\phi}$ have order dividing $e$. Then the exponent of $G$ is $e$-bounded.

Recall that if $T$ is a nilpotent group of class $c$ generated by elements of order dividing $e$, then the exponent of $T$ divides $e^{c}$ (see for example [4, Corollary 2.5.4]). We are now ready to prove Theorem 1.1.

Proof of Theorem 1.1..

In view of Theorem 2.4, it is sufficient to show that $G_{\phi}$ has $(c,d,e)$-bounded exponent. Without loss of generality we can assume that $G=[G,\phi]$. Let $S$ be the set of those elements $h\in G_{\phi}$ for which there exist $x,y\in G_{-\phi}$ such that $h\in\langle x,y\rangle$. Lemma 2.3 says that $G_{\phi}=\langle S\rangle$. Combining Proposition 2.2 with our hypotheses we conclude that the elements in $S$ have $(c,d,e)$-bounded order. Thus, $G_{\phi}$ is a nilpotent group of class $c$ generated by elements of bounded order. Hence, the exponent of $G_{\phi}$ is $\{c,d,e\}$-bounded, as required. ∎

3. Proof of Theorem 1.2

In what follows we write $r(K)$ for the rank of a group $K$. We start this section with a theorem established in [10]. Its proof uses the theory of powerful $p$-groups, due to Lubotzky and Mann [7].

Theorem 3.1**.**

Let $G$ be a finite group of odd order admitting an involutory automorphism $\phi$ such that $r(G_{\phi})=r$. Then $r([G,\phi]^{\prime})$ is $r$-bounded.

Let $p$ be a prime and $K$ a $p$-group. Suppose that a subgroup of $K$ generated by a subset $X\subseteq K$ can be generated by $m$ elements. In the sequel we will use without explicit references the well-known observation that $\langle X\rangle$ can be generated by $m$ elements from the set $X$. This fact can be easily deduced from the Burnside Basis Theorem (see [9, Theorem 5.3.2]). Another well-known fact that we will require is the following lemma.

Lemma 3.2**.**

The order of a finite group $G$ of exponent $e$ and rank $r$ is $(e,r)$-bounded.

Proof.

Indeed, if $G$ is a $p$-group the result is immediate from [5, Corollary 11.21]. Since the order of $G$ is the product of the orders of its Sylow subgroups, the general case follows from the observation that the prime divisors of the order of $G$ are divisors of $e$. ∎

Proof of Theorem 1.2..

Without loss of generality we may assume that $G=[G,\phi]$. Let $r^{\prime}=r(G^{\prime})$. Theorem 3.1 tells us that $r^{\prime}$ is $r$-bounded. It follows that the Fitting height of $G$ is $r$-bounded as well (see for example [6, Lemma 2.4]). Arguing by induction on the Fitting height of $G^{\prime}$ we can assume that the exponent of $G/F(G^{\prime})$ is $(e,r)$-bounded. Thus, there is an $(e,r)$-bounded positive integer $f$ such that $G^{f}\leq F(G^{\prime})$. Set $M=G^{f}$. Note that both the rank and the exponent of $(G/M)_{\phi}$ are $(e,r)$-bounded. We conclude that the order of $(G/M)_{\phi}$ is $(e,r)$-bounded, too (Lemma 3.2). Now a theorem of Hartley and Meixner [3] says that $G/M$ has a subgroup of bounded index which is nilpotent of class at most two. We conclude that the derived length of $G/M$ is $(e,r)$-bounded.

Assume that $M$ is a $p$-group for some prime $p$. Let $A$ be a maximal normal abelian subgroup of $M$. Note that $A=C_{M}(A)$ (see [9, Theorem 5.2.3]). Denote by $B$ the minimal characteristic subgroup of $M$ containing $A$. Since $B$ is a product of at most $r^{\prime}$ subgroups of the form $A^{\alpha}$, where $\alpha\in Aut\,G$, it follows that the nilpotency class of $B$ is at most $r^{\prime}$. Of course, $C_{M}(B)\leq B$. Let $D$ be the minimal normal subgroup of $G$ containing $B_{-\phi}$. This is a subgroup of class at most $r^{\prime}$ generated by elements of order dividing $e$. It follows that the exponent of $D$ is $(e,r)$-bounded. Taking into account that the rank of $D$ is at most $r^{\prime}$ we conclude that the order of $D$ is $(e,r)$-bounded, too. It follows that there is an $(e,r)$-bounded number $j$ such that $D$ is contained in $Z_{j}(M)$, the $j$th term of the upper central series of $M$. Observe that $\phi$ acts trivially on $B/D$. In view of Lemma 2.1.3 we conclude that $B/D\leq Z(G/D)$. Therefore $B$ is contained in $Z_{j+1}(M)$. It follows that the $(j+1)$th term of the lower central series of $M$ centralizes $B$. Recall that $C_{M}(B)\leq B$. Hence the $(j+1)$th term of the lower central series of $M$ is contained in $B$. Taking into account that the nilpotency class of $B$ is at most $r^{\prime}$ we deduce that the derived length of $M$ is $(e,r)$-bounded. Since also the derived length of $G/M$ is $(e,r)$-bounded, it follows that the derived length of $G$ is $(e,r)$-bounded. By Proposition 2.2, the exponent of $G$ is $(e,r)$-bounded.

Thus, the theorem is proved in the particular case where $M$ is a $p$-group. Note that in that case the bound $e_{0}$ on the exponent of $G$ does not depend on the prime $p$. In general $M$ is a direct product of its Sylow $p$-subgroups. The above paragraph shows that the exponent of $G/O_{p^{\prime}}(M)$ divides $e_{0}$ for each prime divisor $p$ of the order of $M$. Since $\bigcap_{p\in\pi(M)}O_{p^{\prime}}(M)=1$, we conclude that the exponent of $G$ divides $e_{0}$. The proof is now complete. ∎