Implications on the Index of a Fixed Point Subgroup 2

Abstract

Let G be a finite group and A be a subgroup of Aut(G). In this work, we study the influence of the index of the fixed-point subgroup of A in G on the structure of G. When A is cyclic, we prove that G is solvable if the index of the centralizer of each x in H-G is square-free, where H denotes the semidirect product of G by A.

1. Introduction

In the present paper, all the groups considered are finite. Solvability of a group is used in determining the algebraic solvability of polynomial equations, advancing cryptography, and solving geometric and topological problems. When we model a real-life problem in which we need solvable groups, we obtain some information about the group but not the complete group. This information may be related to order, conjugacy classes, fixed points of automorphisms, etc. of the group. This allows us to deduce information about the structure of the group by means of some restricted information such as the number of conjugacy classes, the length of conjugacy classes, the number of automorphisms, and the number of fixed points of automorphisms. This makes the problem of finding solvability conditions by means of some given information about the group important. Furthermore, solvability conditions can be converted to conditions for supersolvability and nilpotency, etc., of groups after some more restrictions.

In the literature, there are several types of such solvability criteria. Many of them use arithmetical restrictions on groups. We may separate these kinds of problems into three types: arithmetical conditions on conjugacy classes, arithmetical conditions on the set of fixed points of automorphisms, and the others.

There are several papers using arithmetical conditions on the set of conjugacy classes to determine the solvability of groups. These types of works are categorized and mentioned chronologically in [1]. Also, there are many new results in this area, one of which is [2]. The premises used in these types of problems can be adapted as restrictions on fixed-point subgroups of a group to obtain new criteria for the solvability of groups.

There are several papers using arithmetical conditions on the fixed-point subgroup of automorphisms of a group to classify the structure of the group. Generally, authors consider the influence of the fixed-point subgroup of an automorphism group when it is trivial or small. Examples of such works include [3,4,5]. Also, there are a few results concerning the structure of G when the number of fixed points of an automorphism group is large.

There is an obvious relation between conjugacy classes of a group and fixed-point subgroups of inner automorphisms of the group. This makes us consider arithmetical conditions on the fixed-point subgroups of automorphisms similar to the arithmetical conditions on conjugacy classes considered in the papers related to the problems of the first type. One important result in these types of problems is [6]. Kazarin proved the solvability of $[G,\alpha ]$ where G is a group, $\alpha$ is an automorphism of G whose fixed-point subgroup contains $p^a$ elements for some prime p.

Our aim is to find information about the structure of a group under some chosen arithmetical conditions related to the indices of fixed-point subgroups of automorphisms of the group. In this paper, we consider the influence of fixed-point subgroups of automorphisms with square-free order on the solvability of a group.

2. Preliminaries

Let $\alpha$ be an automorphism of G. We define the semidirect product group $G⋊⟨\alpha ⟩$ by ${\alpha }^{-1}.x.\alpha =\alpha \left(x\right)$ for every $x\in G$ and denote it simply by $G⟨\alpha ⟩$, and elements of this semidirect product are denoted by $g.{\alpha }^k$ for $g\in G$ and $k\in \mathbb{Z}$. By means of this semidirect product, we may view the fixed-point subgroup of $\alpha$ as the centralizer $G_G\left(\alpha \right)$ of $\alpha$ in G and we may denote its index by $In{d}_G\left(\alpha \right)$, similar to the notation used in [7].

In [7], we proved the following theorem for an automorphism whose order is coprime to the order of the group.

Theorem 1

([7], Theorem 1.3). Let G be a finite group and α be an automorphism of G such that $\left(\left|G\right|,\left|\alpha \right|\right)=1$. If $In{d}_G\left(\alpha \right)$ is square-free, then $\left[G,\alpha \right]$ is solvable.

We can see that the conclusion of Theorem 1 is not necessarily valid in the case of a noncoprime action, as demonstrated by the following example.

Example 1.

When $G=A_5$ with $\alpha ={\tau }_{\left(12\right)}\in Aut\left(G\right)\backslash Inn\left(G\right)$ we have $In{d}_G\left(\alpha \right)=10$, but $[A_5,\left(12\right)]=A_5$ is simple.

By supposing some extra conditions on the elements of $G⟨\alpha ⟩\backslash G$, we consider the following problem:

Let G be a finite group and $\alpha \in Aut\left(G\right)$. If $In{d}_G\left(x\right)$ is square-free for each $x\in H\backslash G$ where $H=G⟨\alpha ⟩$, then G is solvable.

In Section 3, we give a proof of the above question and in Section 4, we two nontrivial examples illustrating the above hypothesis.

In order to simplify the proof of the main theorem we present some lemmata and a theorem.

Some of the following lemmata may be obvious results, but we prove them to familiarize the reader with the notation used in this paper. The lemmas presented with proofs are adapted from the author’s thesis work [8], extending arithmetical conditions on conjugacy classes to the fixed-point subgroups of automorphisms.

Lemma 1.

Let G be a group and $\alpha \in Aut\left(G\right)$. For a nontrivial α-invariant proper normal subgroup N of G, if $x\in N⟨\alpha ⟩\backslash N$ then $In{d}_N\left(x\right)$ is a divisor of $In{d}_G\left(x\right)$.

Proof. 

Let $x\in N⟨\alpha ⟩\backslash N$ where N is a nontrivial $\alpha$-invariant proper normal subgroup of G. Then

$$In{d}_N\left(x\right)=\left|N:C_N\left(x\right)\right|=\left|N:N\cap C_G\left(x\right)\right|=\left|C_G\left(x\right)N:C_G\left(x\right)\right|$$

Since N is a normal subgroup of G, we have $C_G\left(x\right)N$ is a subgroup of G and so

$$In{d}_G\left(x\right)=\left|G:C_G\left(x\right)\right|=\left|G:C_G\left(x\right)N\right|.\left|C_G\left(x\right)N:C_G\left(x\right)\right|$$

Therefore, $In{d}_N\left(x\right)$ is a divisor of $In{d}_G\left(x\right)$. □

Let G be a group and $\alpha \in Aut\left(G\right)$. For an $\alpha$-invariant proper normal subgroup N of G, $\alpha$ induces the automorphism $\overline{\alpha }:G/N\to G/N$ defined for every $xN\in G/N$ by $\overline{\alpha }\left(xN\right)=\alpha \left(x\right)N$.

The map

$$\varphi :G⟨\alpha ⟩/N\to \left(G/N\right)⟨\overline{\alpha }⟩$$

given by

$$\varphi (x.{\alpha }^{k}N)=\left(xN\right).{\overline{\alpha }}^k$$

for $x\in G$ and $k\in \mathbb{Z}$ is an isomorphism.

Lemma 2.

Let G be a group and $\alpha \in Aut\left(G\right)$. For a nontrivial α-invariant proper normal subgroup N of G, if $y\in \left(G/N\right)⟨\overline{\alpha }⟩\backslash \left(G/N\right)$ then $In{d}_{G/N}\left(y\right)$ is a divisor of $In{d}_G(x.{\alpha }^k)$ where $x\in G$ and $k\in \mathbb{Z}$ so that $y=\left(xN\right).{\overline{\alpha }}^k$.

Proof. 

Let $y\in \left(G/N\right)⟨\overline{\alpha }⟩\backslash \left(G/N\right)$ where N is a nontrivial $\alpha$-invariant proper normal subgroup of G. Then there are $x\in G$ and $k\in \mathbb{Z}$ so that $y=\left(xN\right).{\overline{\alpha }}^k$. It follows that

$$In{d}_{G/N}\left(y\right)=In{d}_{G/N}\left(\left(xN\right).{\overline{\alpha }}^k\right)=\left|\left(G/N\right):C_{\left(G/N\right)}(\left(xN\right).{\overline{\alpha }}^k)\right|$$

is a divisor of

$$\left|\left(G/N\right):C_G(x.{\alpha }^k)N/N\right|=\left|G:C_G(x.{\alpha }^k)N\right|$$

Since N is a normal subgroup of G, we have $C_G(x.{\alpha }^k)N$ is a subgroup of G and so

$$In{d}_G(x.{\alpha }^k)=\left|G:C_G(x.{\alpha }^k)\right|=\left|G:C_G(x.{\alpha }^k)N\right|.\left|C_G(x.{\alpha }^k)N:C_G\left(x\right)\right|$$

Therefore, $In{d}_{G/N}\left(y\right)$ is a divisor of $In{d}_G(x.{\alpha }^k)$. □

Lemma 3.

Let G be a group and $g\in G$. If $x\in G⟨{\tau }_g⟩\backslash G$ where ${\tau }_g$ is the inner automorphism of G induced by g defined by ${\tau }_g\left(a\right)=g^{-1}ag$ for all $a\in G$. Then $C_G\left(x{g}^k\right)=C_G(x.{\tau }_g^k)$ for all $x\in G$ and $k\in \mathbb{Z}$.

Proof. 

Let $x\in G$ and $k\in \mathbb{Z}$.

Since

$$b^{x.{\tau }_g^k}={\tau }_g^{-k}.x^{-1}bx.{\tau }_g^k=g^{-k}{x}^{-1}bx{g}^k=b^{x{g}^k}$$

we observe that

$$C_G\left(x{g}^k\right)=C_G(x.{\tau }_g^k)$$

□

Lemma 4.

For $n\ge 5$ for an involution $\rho \in S_n\backslash A_n$, $In{d}_{A_n}(x.{\tau }_{\rho })$ is divisible by 4 for some $x\in A_n$.

Proof. 

Let $\delta \in S_n\backslash A_n$ be an involution. Then ${\tau }_{\delta }$ is an inner automorphism of $S_n$ induced by $\delta$. By Lemma 3, for every $h\in S_n$ we have

$$C_{S_n}\left(h\delta \right)=C_{S_n}(h.{\tau }_{\delta })$$

Let $\rho =\left(12\right)$.

If n is odd, let h be an $(n-2)$-cycle fixing 1 and 2. Then $h\in A_n$ and $h\rho$ is a $\{2,n-2\}$-cycle and $\left|C_{S_n}\left(h\rho \right)\right|=2(n-2)$. Since $\rho \in C_{S_n}\left(h\rho \right)$ we get

$$\left|C_{A_n}(h.{\tau }_{\rho })\right|=\left|C_{A_n}\left(h\rho \right)\right|=(n-2).$$

It follows that $In{d}_{A_n}(h.{\tau }_{\rho })={\displaystyle \frac{n!}{2(n-2)}}$ is divisible by 4.

If n is even, let h be an $(n-3)$-cycle fixing 1 and 2. Then $h\in A_n$ and $h\rho$ is a $\{2,n-3\}$-cycle and $\left|C_{S_n}\left(h\rho \right)\right|=2(n-3)$. Since $\rho \in C_{S_n}\left(h\rho \right)$ we get

$$\left|C_{A_n}(h.{\tau }_{\rho })\right|=\left|C_{A_n}\left(h\rho \right)\right|=(n-3).$$

It follows that $In{d}_{A_n}(h.{\tau }_{\rho })={\displaystyle \frac{n!}{2(n-3)}}$ is divisible by 4.

Let $\sigma \in S_n\backslash A_n$ be an arbitrary involution. Then $\rho {\sigma }^{-1}\in A_n$. It follows that for the chosen $h\in A_n$ above we have

$$C_{S_n}(h.{\tau }_{\rho })=C_{S_n}\left(h\rho \right)=C_{S_n}\left(h\rho {\sigma }^{-1}\sigma \right)=C_{S_n}(h\rho {\sigma }^{-1}.{\tau }_{\sigma })$$

So, $x=h\rho {\sigma }^{-1}\in A_n$ and $In{d}_{A_n}(x.{\tau }_{\sigma })=In{d}_{A_n}(h.{\tau }_{\rho })$ is divisible by 4.

Therefore, for an involution $\sigma \in S_n\backslash A_n$, there is $x\in A_n$ so that $In{d}_{A_n}(x.{\tau }_{\sigma })$ is divisible by 4. □

Lemma 5.

Let G be a group and x and y be two commuting elements of G such that $(\left|x\right|,\left|y\right|)=1$. Then $C_G\left(xy\right)=C_G\left(x\right)\cap C_G\left(y\right)$.

Proof. 

Trivially, $C_G\left(x\right)\cap C_G\left(y\right)\le C_G\left(xy\right)$.

If $\left|x\right|=m$ and $\left|y\right|=n$. Since $(m,n)=1$, there exist integers $r,s$ such that $rm+sn=1$. As $[x,y]=1$, we have

$${\left(xy\right)}^{rm}=x^{rm}{y}^{rm}=y^{1-sn}=y$$

and

$${\left(xy\right)}^{sn}=x^{sn}{y}^{sn}=x^{1-rm}=x.$$

If $a\in C_G\left(xy\right)$, then

$$1=[a,{\left(xy\right)}^{rm}]=[a,y]$$

and

$$1=[a,{\left(xy\right)}^{sn}]=[a,x].$$

It follows that $a\in C_G\left(x\right)$ and $a\in C_G\left(y\right)$. Therefore,

$$C_G\left(xy\right)=C_G\left(x\right)\cap C_G\left(y\right)$$

as desired. □

Lemma 6

([9], Lemma 6). If G is simple, then 4 divides the length of a conjugacy class of G.

Lemma 7

([10], Lemma 1, Corollary 1). Let G be a finite group.

1. 

If for some prime number p, any $p^{\prime }$-element has $p^{\prime }$-index in G, then G has a unique Sylow p-subgroup which is a direct factor.

2. 

If for some prime number p, $p\nmid \left|C\right|$ for any $C\in Con\left(G\right)$ then G has a unique Sylow p-subgroup which is an abelian direct factor.

Theorem 2

([11], Theorem B). Let G be a finite simple non-abelian group and let p be an odd prime dividing the order of G. Let $A=Aut\left(G\right)$ and let S be a Sylow p-subgroup of A. Identifying G with $Inn\left(G\right)$, set $P=S\cap G$. Then $C_S\left(P\right)=Z\left(P\right)$.

3. Main Theorem

We prove the following theorem as a result in my thesis work ([8], Theorem 3):

Theorem 3.

Let G be a finite group and $\alpha \in Aut\left(G\right)$. If $In{d}_G\left(x\right)$ is square-free for each $x\in H\backslash G$ where $H=G⟨\alpha ⟩$, then G is solvable.

Proof. 

The proof of the theorem is based on obtaining a contradiction over a series of steps by induction on the order of the semidirect product group H. Let G be a group and $\alpha$ be an automorphism of G and $H=G⟨\alpha ⟩$ be an ordered triple such that H is of the smallest order among such triples satisfying premises of the theorem but G is not solvable.

• G has no nontrivial proper normal $\alpha$-invariant subgroup. In particular, $G=[G,\alpha ]$.

Assume that N is a nontrivial $\alpha$-invariant proper normal subgroup of G.

Let $x\in N⟨\alpha ⟩\backslash N$. Then $x\in H\backslash G$ and so $In{d}_G\left(x\right)$ is square-free. Since $In{d}_N\left(x\right)$ is a divisor of $In{d}_G\left(x\right)$ by Lemma 1, we obtain $In{d}_N\left(x\right)$ is also square-free. As $\left|N⟨\alpha ⟩\right|<\left|H\right|$, N is solvable by induction assumption.

Let $y\in \left(G/N\right)⟨\alpha ⟩\backslash \left(G/N\right)$. Then $y=\left(xN\right){\alpha }^k$ for some $x\in G$ and for some integer k. It follows by Lemma 2 that, $In{d}_{G/N}\left(y\right)$ is a divisor of $In{d}_G\left(x{\alpha }^k\right)$. Since $x{\alpha }^k\in H\backslash G$, we have $In{d}_G\left(x{\alpha }^k\right)$ is square-free. Hence, $In{d}_{G/N}\left(y\right)$ is also square-free. Hence, by the induction hypothesis, $G/N$ is solvable as $\left|\left(G/N\right)⟨\alpha ⟩\right|<\left|H\right|$. This forces G to be solvable, which contradicts our assumption.

Therefore, G cannot have a nontrivial proper normal $\alpha$-invariant subgroup.

In particular, since $[G,\alpha ]$ is an $\alpha$-invariant subgroup of G, we obtain $[G,\alpha ]=G$.

2.

The order of $\alpha$ is a prime p, and p divides $\left|G\right|$.

Let $\left|\alpha \right|=kp$ for some prime p and a positive integer k. Then the order of ${\alpha }^k$ is p. Let $x\in G⟨{\alpha }^k⟩\backslash G$. It follows that $In{d}_G\left(x\right)$ is square-free as $x\in H\backslash G$. As $\left|G⟨{\alpha }^k⟩\right|<\left|H\right|$, by induction assumption we get G is solvable. This contradicts our assumption if $k>1$. Hence, the order of $\alpha$ is the prime number p. Theorem 1 implies that p divides $\left|G\right|$ since $[G,\alpha ]=G$.

3.

G is simple.

G is characteristically simple by Step 1. That is,

$$G=E_1\times \dots \times E_s$$

where each $E_i$, $i=1,\dots ,s$, is a nonabelian simple group isomorphic to $E_1$. Suppose $s>1$. Let $\{E_1,E_1^{\alpha },\dots ,E_1^{{\alpha }^n}\}$ be the orbit containing $E_1$ under the action of $\alpha$ on the subgroups of G. Since the order of $\alpha$ is the prime number p, the length $n\ne 1$ of the orbit is a divisor of p. So $n=p$. Then

$$C=E_1\times E_1^{\alpha }\times \dots \times E_1^{{\alpha }^n}$$

is an $\alpha$-invariant normal subgroup of G and hence $G=C$ by Step 1. Now, $\left|C_G\left(\alpha \right)\right|=\left|E_1\right|$ and we have $In{d}_G\left(\alpha \right)={\left|E_1\right|}^{s-1}$ is square-free. This is impossible as 4 divides $\left|E_1\right|$ by Lemma 6. Therefore $s=1$, i.e., G is a nonabelian simple group.

4.

$\alpha$ is not an inner automorphism.

Suppose that $\alpha$ is an inner automorphism. Then there is $g\in G$ so that $\alpha ={\tau }_g$. Since G is simple by step 3, there is $a\in G$ so that $In{d}_G\left(a\right)$ is a multiple of 4 by Lemma 6. It follows by Lemma 3 that,

$$C_G\left(a\right)=C_G\left(a{g}^{-1}g\right)=C_G(a{g}^{-1}.{\tau }_g).$$

Hence, $In{d}_G(a{g}^{-1}.{\tau }_g)=In{d}_G\left(a\right)$ is divisible by 4 and since $a{g}^{-1}\in G$ and $\alpha ={\tau }_g$, we have $a{g}^{-1}.{\tau }_g\in H\backslash G$. This contradicts our assumption.

Therefore, $\alpha$ cannot be an inner automorphism.

5.

G is not an alternating group:

Suppose that G is an alternating group. Since G is simple, we have $G=A_n$ where $n\ge 5$. We can divide the proof of this step into two parts.

Let $n\ne 6$. We know that $\mathrm{Aut}\left(A_n\right)\cong S_n$ and every automorphism of $A_n$ is the restriction of an inner automorphism ${\tau }_{\rho }$ of $S_n$ to $A_n$ for some $\rho \in S_n$. Since $\alpha$ is not an inner automorphism and is of prime order, it must be an automorphism of order 2, which is the restriction of ${\tau }_{\rho }$ to $A_n$ for some involution $\rho \in S_n\backslash A_n$. By Lemma 4, there is an element $x\in A_n$ so that $In{d}_{A_n}(x.{\tau }_{\rho })$ is divisible by 4, which contradicts our assumption.

Let $n=6$. Since $\alpha$ is an automorphism of $A_6$ which is not an inner automorphism and is of prime order, it is induced by either an inner automorphism of $S_6$ or an outer automorphism of $S_6$ sending 3-cycles to $\{3,3\}$-cycles and $\{3,3\}$-cycles to 3-cycles ([12], Section 2.4). The former case is not possible by Lemma 4. So $\alpha$ is induced by an outer automorphism of $S_n$. Any 3-element of G is either a 3-cycle or a $\{3,3\}$-cycle. If 3 divides $\left|C_G\left(\alpha \right)\right|$, then there will be a 3-cyle or a $\{3,3\}$-cyle which is fixed by $\alpha$ which contradicts the definition of $\alpha$. Thus, $In{d}_G\left(\alpha \right)$ is divisible by 9, which is not the case.

Therefore, G is not an alternating group.

6.

G is not a sporadic group.

By using the Atlas of Finite Groups [13], only

$$M_{12},M_{22},J_2,J_3,F{i}_{22},F{i}_{24},HS,M^{c}L,He,Suz,O^{\prime }N,HN$$

have noninner automorphism.

If G is one of these groups, since $\left|\mathrm{Aut}\left(G\right):\mathrm{Inn}\left(G\right)\right|=2$ and $\alpha$ is a noninner automorphism of G of prime order, then order of $\alpha$ should be 2 and $G⟨\alpha ⟩\cong \mathrm{Aut}\left(G\right)$. Then, from character table, we see that some conjugacy classes of $\mathrm{Aut}\left(G\right)\backslash G$ have length divisible by 4.

Therefore, G is not a sporadic group. As a conclusion G is a simple group of Lie Type.

7.

$In{d}_G\left(\alpha \right)$ is divisible by p when p is odd.

Assume that p is odd. If p does not divide $In{d}_G\left(\alpha \right)$, then $\alpha$ centralizes a Sylow p-subgroup P of G. It follows from Theorem 9 that $\alpha$ is induced by an element of $Z\left(P\right)$, which is impossible by Step 4.

8.

We may assume that $p=2$.

Suppose that p is odd. Then by step 7, p divides $In{d}_G\left(\alpha \right)$. If y is a $p^{\prime }$-element of $C_G\left(\alpha \right)$, then by Lemma 5, we have

$$C_{C_G\left(\alpha \right)}\left(y\right)=C_G\left(\alpha \right)\cap C_G\left(y\right)=C_G\left(y\alpha \right)\subseteq C_G\left(\alpha \right)$$

Now,

$$\begin{array}{ccc}In{d}_G\left(y\alpha \right)& =& \left|G:C_G\left(y\alpha \right)\right|\\ & =& \left|G:C_G\left(\alpha \right)\right|\left|C_G\left(\alpha \right):C_{C_G\left(\alpha \right)}\left(y\right)\right|\\ & =& In{d}_G\left(\alpha \right)\left|C_G\left(\alpha \right):C_{C_G\left(\alpha \right)}\left(y\right)\right|\end{array}$$

is square-free and hence $p\nmid \left|C_G\left(\alpha \right):C_{C_G\left(\alpha \right)}\left(y\right)\right|$.

It follows by Lemma 7 (1) that, $C_G\left(\alpha \right)$ has a unique Sylow p-subgroup P which is a direct factor, i.e., $C_G\left(\alpha \right)=P\times A$ where A is a $p^{\prime }$-group. Let $a\in A$. Now, $C_{C_G\left(\alpha \right)}\left(a\right)=P\times C_A\left(a\right)$ and also $C_{C_G\left(\alpha \right)}\left(a\right)=C_G\left(\alpha \right)\cap C_G\left(a\right)=C_G\left(a\alpha \right)\subseteq C_G\left(\alpha \right)$ by Lemma 5. So $\left|C_G\left(\alpha \right):C_{C_G\left(\alpha \right)}\left(a\right)\right|=\left|P\times A:P\times C_A\left(a\right)\right|=\left|A:C_A\left(a\right)\right|=In{d}_A\left(a\right)$ is a divisor of $In{d}_G\left(a\alpha \right)$ and hence $In{d}_A\left(a\right)$ is square-free.

If $b\in A$ is an involution, then for any $x\in A⟨{\tau }_b⟩\backslash A$, there exists $c\in A$ such that $x=c{\tau }_b$ and hence $C_A\left(cb\right)=C_A\left(x\right)$ by Lemma 3. Therefore, $In{d}_A\left(x\right)$ is square-free. As p is odd, we have $\left|A⟨{\tau }_b⟩\right|<\left|H\right|$. It follows by induction that A is solvable. This forces the solvability of $C_G\left(\alpha \right)$ if it contains a $p^{\prime }$-element. In case $C_G\left(\alpha \right)$ is a p-group, it is already solvable.

When G is a simple group of Lie type and $\alpha$ is a noninner automorphism of odd prime order, then $C_G\left(\alpha \right)$ has a nonabelian simple subgroup by [14], Theorem 4.2.2 when $\alpha$ is an inner-diagonal or graph automorphism when p is coprime to characteristic r of the base field, and by [14], Proposition 4.9.1 when $\alpha$ is a field automorphism, and by [14], Proposition 4.9.2 when $p=3$, $\alpha$ is a graph automorphism of a simple group of Lie type over a field with characteristic 3. This contradicts the solvability of $C_G\left(\alpha \right)$.

Therefore, we may assume that $p=2$.

9.

Final Contradiction.

We consider three cases:

By [15] Chapter 17, Lemma 7.2, if G is a simple group of Lie Type over a field of odd characteristic, then every involution $\alpha \in Aut\left(G\right)\backslash Inn\left(G\right)$ maps into $Outdiag\left(G\right)$ or $\Phi$ i.e., $\alpha$ is either an inner-diagonal or a graph or a field or a graph-field automorphism. We consider inner-diagonal and graph automorphisms for such groups in case 1. In case 2, we study field and graph-field automorphisms where there is no restriction on the characteristic r of the base field. In characteristic 2, there is no nontrivial diagonal automorphism. So $\alpha$ is either a graph or a field or a graph-field automorphism since both $\alpha$ is of order 2 and field automorphisms commute with graph automorphisms. Since field and graph-field automorphisms are considered for every characteristic in case 3, we studied the remaining automorphisms, i.e., graph automorphisms of simple groups of Lie type over a field of characteristic 2.

Let the largest normal subgroup of $C_G\left(\alpha \right)$ with $r^{\prime }$ index in $C_G\left(\alpha \right)$ be denoted by $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$. If $|C_G\left(\alpha \right)|=r^{a}b$, then $r^a$ is also a divisor of order of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ since $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ contains a Sylow r-subgroup of $C_G\left(\alpha \right)$, because $\left|C_G\left(\alpha \right):O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)\right|$ is an $r^{\prime }$ number. Hence, the power of r in the prime decomposition of both $In{d}_G\alpha$ and $\left|G:O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)\right|$ are the same. Let ${\left|n\right|}_r$ denote the r-power part of the prime decomposition of the positive integer n.

Case 1. $\alpha$ is an inner-diagonal or graph automorphism where q is odd.

Tables 4.5.1, 4.5.2, and 4.5.3 in [14] give all possible inner-diagonal or graph automorphisms of simple groups of Lie type over a field of odd characteristic and the largest normal $r^{\prime }$-indexed subgroup structure of their fixed-point subgroups.

We consider all the families of simple groups of Lie type and calculated

$$|In{d}_G{\left(\alpha \right)|}_r={\left|G:O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)\right|}_r={\displaystyle \frac{{\left|G\right|}_r}{{\left|O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)\right|}_r}}$$

for each case.

(i)    

$G=A_m^{\epsilon }\left(q\right)$ for $m\ge 2$, $\epsilon =\pm 1$.

From Table 4.5.2 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

${\displaystyle P_1=A_{m-1}^{\epsilon }\left(q\right)}$,

${\displaystyle P_2=A_{i-1}^{\epsilon }\left(q\right)A_{m-i}^{\epsilon }\left(q\right)}$ and $2\le i\le {\displaystyle \frac{m}{2}}$,

${\displaystyle P_3=A_{\frac{m-1}{2}}^{\epsilon }{\left(q\right)}^2}$ and m is odd,

${\displaystyle P_4=A_{\frac{m-1}{2}}^{\epsilon }\left(q^2\right)}$ and m is odd,

${\displaystyle P_5=C_{\frac{m+1}{2}}\left(q\right)}$ and m is odd,

${\displaystyle P_6=B_{\frac{m}{2}}\left(q\right)}$ and m is even,

${\displaystyle P_7=D_{\frac{m+1}{2}}\left(q\right)}$ and m is odd,

${\displaystyle P_8{=}^2{D}_{\frac{m+1}{2}}\left(q\right)}$ and m is odd.

We have

$${\left|G:P_1\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{\frac{(m-1)m}{2}}}}=q^m$$

$${\left|G:P_2\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{\frac{(i-1)i}{2}}{q}^{\frac{(m-i)(m-i+1)}{2}}}}=q^{mi-i^2}$$

$${\left|G:P_3\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{2\frac{\frac{m-1}{2}\frac{m+1}{2}}{2}}}}=q^{{\displaystyle \frac{{(m+1)}^2}{4}}}$$

$${\left|G:P_4\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{2\frac{\frac{m-1}{2}\frac{m+1}{2}}{2}}}}=q^{{\displaystyle \frac{{(m+1)}^2}{4}}}$$

$${\left|G:P_5\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{{\left(\frac{m+1}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2-1}{4}}}$$

$${\left|G:P_6\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{{\left(\frac{m}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2+2m}{4}}}$$

$${\left|G:P_7\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{\frac{m+1}{2}\frac{m-1}{2}}}}=q^{{\displaystyle \frac{{(m+1)}^2}{4}}}$$

$${\left|G:P_8\right|}_r={\displaystyle \frac{q^{\frac{m(m+1)}{2}}}{q^{\frac{m+1}{2}\frac{m-1}{2}}}}=q^{{\displaystyle \frac{{(m+1)}^2}{4}}}$$

Since $m\ge 2$, in each case $r^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

(ii)   

$G=B_m\left(q\right)$ for $m\ge 2$.

From Table 4.5.2 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

$Q_1=B_{m-1}\left(q\right)$,

$Q_2=B_{m-1}\left(q\right)$,

$Q_3=D_i\left(q\right)B_{m-i}\left(q\right)$ and $2\le i<m$,

$Q_4{=}^2{D}_i\left(q\right)B_{m-i}\left(q\right)$ and $2\le i<m$,

$Q_5=D_m\left(q\right)$,

$Q_6{=}^2{D}_m\left(q\right)$.

We have

$${\left|G:Q_1\right|}_r={\displaystyle \frac{q^{m^2}}{q^{{(m-1)}^2}}}=q^{2m-1}$$

$${\left|G:Q_2\right|}_r={\displaystyle \frac{q^{m^2}}{q^{{(m-1)}^2}}}=q^{2m-1}$$

$${\left|G:Q_3\right|}_r={\displaystyle \frac{q^{m^2}}{q^{i(i-1)}{q}^{{(m-i)}^2}}}=q^{2mi+i-2i^2}$$

$${\left|G:Q_4\right|}_r={\displaystyle \frac{q^{m^2}}{q^{i(i-1)}{q}^{{(m-i)}^2}}}=q^{2mi+i-2i^2}$$

$${\left|G:Q_5\right|}_r={\displaystyle \frac{q^{m^2}}{q^{m(m-1)}}}=q^m$$

$${\left|G:Q_6\right|}_r={\displaystyle \frac{q^{m^2}}{q^{m(m-1)}}}=q^m$$

Since $m\ge 2$, in each case $r^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

(iii)  

$G=C_m\left(q\right)$ for $m\ge 2$.

From Table 4.5.2 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

$R_1=C_i\left(q\right)C_{m-i}\left(q\right)$ and $1\le i<{\displaystyle \frac{m}{2}}$,

$R_2=C_{{\displaystyle \frac{m}{2}}}{\left(q\right)}^2$ and m is even,

$R_3=C_{{\displaystyle \frac{m}{2}}}\left(q^2\right)$ and m is even,

$R_4=A_{m-1}\left(q\right)$,

$R_5{=}^2{A}_{m-1}\left(q\right)$.

We have

$${\left|G:R_1\right|}_r={\displaystyle \frac{q^{m^2}}{q^{i^2}{q}^{{(m-i)}^2}}}=q^{2mi-2i^2}$$

$${\left|G:R_2\right|}_r={\displaystyle \frac{q^{m^2}}{q^{2{\left(\frac{m}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2}{2}}}$$

$${\left|G:R_3\right|}_r={\displaystyle \frac{q^{m^2}}{q^{2{\left(\frac{m}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2}{2}}}$$

$${\left|G:R_4\right|}_r={\displaystyle \frac{q^{m^2}}{q^{\frac{m(m-1)}{2}}}}=q^{{\displaystyle \frac{m^2+m}{2}}}$$

$${\left|G:R_5\right|}_r={\displaystyle \frac{q^{m^2}}{q^{\frac{m(m-1)}{2}}}}=q^{{\displaystyle \frac{m^2+m}{2}}}$$

Since $m\ge 2$, in each case $r^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

(iv)  

$G=D_m^{\epsilon }\left(q\right)$ for $m\ge 4$, $\epsilon =\pm 1$.

From Tables 4.5.2 and 4.5.3 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

$S_1=D_{m-1}^{\epsilon }\left(q\right)$,

$S_2=D_{m-1}^{-\epsilon }\left(q\right)$,

$S_3=D_i\left(q\right)D_{m-i}^{\epsilon }\left(q\right)$ and $2\le i<{\displaystyle \frac{m}{2}}$,

$S_4{=}^2{D}_i\left(q\right)D_{m-i}^{-\epsilon }\left(q\right)$ and $2\le i<{\displaystyle \frac{m}{2}}$,

$S_5=D_{{\displaystyle \frac{m}{2}}}{\left(q\right)}^2$ and $G=D_{2k}^{+}\left(q\right)$,

$S_6{=}^2{D}_{{\displaystyle \frac{m}{2}}}{\left(q\right)}^2$ and $G=D_{2k}^{+}\left(q\right)$,

$S_7=A_{m-1}\left(q\right)$ and $G=D_{2k}^{+}\left(q\right)$,

$S_8{=}^2{A}_{m-1}\left(q\right)$ and $G=D_{2k}^{+}\left(q\right)$,

$S_9{=}^2{D}_{{\displaystyle \frac{m}{2}}}\left(q\right)D_{{\displaystyle \frac{m}{2}}}\left(q\right)$ and $G=D_{2k}^{-}\left(q\right)$,

$S_{10}{=}^2{D}_{{\displaystyle \frac{m}{2}}}\left(q^2\right)$ and $G=D_{2k}^{-}\left(q\right)$,

$S_{11}=A_{m-1}^{\epsilon }\left(q\right)$ and $G=D_{2k+1}^{\epsilon }\left(q\right)$,

$S_{12}=B_{m-1}\left(q\right)$,

$S_{13}=B_{i-1}\left(q\right)B_{m-i}\left(q\right)$ and $2\le i\le {\displaystyle \frac{m}{2}}$,

$S_{14}=B_{{\displaystyle \frac{m-1}{2}}}{\left(q\right)}^2$ and $G=D_{2k+1}^{\epsilon }\left(q\right)$,

$S_{15}=B_{{\displaystyle \frac{m-1}{2}}}\left(q^2\right)$ and $G=D_{2k+1}^{\epsilon }\left(q\right)$,

$S_{16}=B_3\left(q\right)$ and $G=D_4^{+}\left(q\right)$,

$S_{17}=A_1\left(q\right)B_2\left(q\right)$ and $G=D_4^{+}\left(q\right)$.

We have

$${\left|G:S_1\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{(m-1)(m-2)}}}=q^{2m-2}$$

$${\left|G:S_2\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{(m-1)(m-2)}}}=q^{2m-2}$$

$${\left|G:S_3\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{i(i-1)}{q}^{(m-i)(m-i-1)}}}=q^{2mi-2i^2}$$

$${\left|G:S_4\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{i(i-1)}{q}^{(m-i)(m-i-1)}}}=q^{2mi-2i^2}$$

$${\left|G:S_5\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2\frac{m}{2}\frac{m-2}{2}}}}=q^{{\displaystyle \frac{m^2-2m+2}{2}}}$$

$${\left|G:S_6\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2\frac{m}{2}\frac{m-2}{2}}}}=q^{{\displaystyle \frac{m^2-2m+2}{2}}}$$

$${\left|G:S_7\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{\frac{m(m-1)}{2}}}}=q^{{\displaystyle \frac{m^2-m}{2}}}$$

$${\left|G:S_8\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{\frac{m(m-1)}{2}}}}=q^{{\displaystyle \frac{m^2-m}{2}}}$$

$${\left|G:S_9\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2\frac{m}{2}\frac{m-2}{2}}}}=q^{{\displaystyle \frac{m^2-2m+2}{2}}}$$

$${\left|G:S_{10}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2\frac{m}{2}\frac{m-2}{2}}}}=q^{{\displaystyle \frac{m^2-2m+2}{2}}}$$

$${\left|G:S_{11}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{\frac{m(m-1)}{2}}}}=q^{{\displaystyle \frac{m^2-m}{2}}}$$

$${\left|G:S_{12}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{{(m-1)}^2}}}=q^{m-1}$$

$${\left|G:S_{13}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{{(i-1)}^2}{q}^{{(m-i)}^2}}}=q^{2mi-i^2+2i-m-1}$$

$${\left|G:S_{14}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2{\left(\frac{m-1}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2-1}{2}}}$$

$${\left|G:S_{15}\right|}_r={\displaystyle \frac{q^{m(m-1)}}{q^{2{\left(\frac{m-1}{2}\right)}^2}}}=q^{{\displaystyle \frac{m^2-1}{2}}}$$

$${\left|G:S_{16}\right|}_r={\displaystyle \frac{q^{4.3}}{q^9}}=q^3$$

$${\left|G:S_{17}\right|}_r={\displaystyle \frac{q^{4.3}}{q^5}}=q^7$$

Since $m\ge 4$, in each case $r^4$ is a divisor of $In{d}_G\left(\alpha \right)$.

(v)   

$G=E_6^{\epsilon }\left(q\right)$ for $\epsilon =\pm 1$.

From Table 4.5.2 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

$T_1=D_5^{\epsilon }\left(q\right)$,

$T_2=A_1\left(q\right)A_5^{\epsilon }\left(q\right)$,

$T_3=F_4\left(q\right)$,

$T_4=C_4\left(q\right)$.

We have

$${\left|G:T_1\right|}_r={\displaystyle \frac{q^{36}}{q^{20}}}=q^{16}$$

$${\left|G:T_2\right|}_r={\displaystyle \frac{q^{36}}{q^{16}}}=q^{20}$$

$${\left|G:T_3\right|}_r={\displaystyle \frac{q^{36}}{q^{24}}}=q^{12}$$

$${\left|T:P_4\right|}_r={\displaystyle \frac{q^{36}}{q^{16}}}=q^{20}$$

In each case $r^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

(vi)  

$G=E_7\left(q\right)$.

From Table 4.5.2 in [14], all the possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are:

$U_1=A_1\left(q\right)D_6\left(q\right)$,

$U_2=A_7\left(q\right)$,

$U_3{=}^2{A}_7\left(q\right)$,

$U_4=E_6\left(q\right)$,

$U_5{=}^2{E}_6\left(q\right)$.

We have,

$${\left|G:U_1\right|}_r={\displaystyle \frac{q^{63}}{q^{31}}}=q^{32}$$

$${\left|G:U_2\right|}_r={\displaystyle \frac{q^{63}}{q^{28}}}=q^{35}$$

$${\left|T:U_3\right|}_r={\displaystyle \frac{q^{63}}{q^{28}}}=q^{35}$$

$${\left|G:U_4\right|}_r={\displaystyle \frac{q^{63}}{q^{36}}}=q^{27}$$

$${\left|G:U_5\right|}_r={\displaystyle \frac{q^{63}}{q^{36}}}=q^{27}$$

In each case $r^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

(vii) 

$G=3_4^D\left(q\right)$

From Table 4.5.1 in [14], only possible structure of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ is $V=A_1\left(q\right)A_1\left(q^3\right)$.

We have

$${\left|G:V\right|}_r={\displaystyle \frac{q^{12}}{q^4}}=q^8$$

It is divisible by $r^2$.

(viii)

$G=G_2\left(q\right)$

From Table 4.5.1 in [14], only possible structure of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ is $W=A_1\left(q^2\right)$.

We have

$${\left|G:W\right|}_r={\displaystyle \frac{q^6}{q^2}}=q^4$$

It is divisible by $r^2$.

(ix)  

$G{=}^2{G}_2\left(q\right)$

From Table 4.5.1 in [14], only possible structure of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ is $Y=A_1\left(q^2\right)$.

We have

$${\left|G:Y\right|}_r={\displaystyle \frac{q^3}{q^2}}=q=3^{2n+1}$$

It is divisible by $r^2$.

(x)   

$F_4\left(q\right)$

From Table 4.5.1 in [14], two possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are

$Z_1=A_1\left(q\right)C_3\left(q\right)$,

$Z_2=B_4\left(q\right)$.

We have

$${\left|G:Z_1\right|}_r={\displaystyle \frac{q^{24}}{q^{10}}}=q^{14}$$

$${\left|G:Z_2\right|}_r={\displaystyle \frac{q^{24}}{q^{16}}}=q^8$$

Both are divisible by $r^2$.

(xi)  

$E_8\left(q\right)$

From Table 4.5.1 in [14], two possible structures of $O^{r^{\prime }}\left(C_G\left(\alpha \right)\right)$ are

$H_1=D_8\left(q\right)$,

$H_2=A_1\left(q\right)E_7\left(q\right)$.

We have

$${\left|G:H_1\right|}_r={\displaystyle \frac{q^{120}}{q^{56}}}=q^{64}$$

$${\left|G:H_2\right|}_r={\displaystyle \frac{q^{120}}{q^{63}}}=q^{57}$$

Both are divisible by $r^2$.

Therefore, in each subcase, $r^2$ divides $\left|In{d}_G\left(\alpha \right)\right|$.

Case 2: $\alpha$ is a field automorphism or a graph-field automorphism.

By Proposition 4.9.1 in [14], if $q=r^{2s}$ for some integer s and $G=L\left(q\right)$, then $C_G\left(\alpha \right)\cong L\left(r^s\right)$.

We consider ${\left|In{d}_G\left(\alpha \right)\right|}_r$ for each family of simple groups of Lie type.

(i)    

If $G\cong A_m\left(r^{2s}\right)$ for $m\ge 2$ then

$$C_G\left(\alpha \right)\cong A_m\left(r^s\right){\mathrm{or}}^2{A}_m\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{{\displaystyle \frac{sm(m+1)}{2}}}.$$

(ii)   

If $G\cong B_m\left(r^{2s}\right)$ for $m\ge 2$ then

$$C_G\left(\alpha \right)\cong B_m\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{s{m}^2}$$

(iii)  

If $G\cong C_m\left(q^{2s}\right)$ for $m\ge 3$ then

$$C_G\left(\alpha \right)\cong C_m\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{s{m}^2}$$

(iv)  

If $G\cong D_m\left(q^{2s}\right)$ for $m\ge 4$ then

$$C_G\left(\alpha \right)\cong D_m\left(r^s\right){\mathrm{or}}^2{D}_m\left(r^s\right){\mathrm{or}}^3{D}_4\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{sm(m-1)}\mathrm{or}{r}^{12s}$$

(v)   

If $G\cong E_6\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right)\cong E_6\left(r^s\right){\mathrm{or}}^2{E}_6\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{36s}$$

(vi)  

If $G\cong E_7\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right)\cong E_7\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{63s}$$

(vii) 

If $G\cong E_8\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right)\cong E_8\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{120s}$$

(viii)

If $G\cong F_4\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right)\cong F_4\left(r^s\right){\mathrm{or}}^2{F}_4\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{24s}\mathrm{or}{r}^{36s}$$

(ix)  

If $G\cong G_2\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right)\cong G_2\left(r^s\right){\mathrm{or}}^2{G}_2\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{6s}\mathrm{or}{r}^{9s}$$

(x)   

If $G{\cong }^2{A}_m\left(r^{2s}\right)$ for $m\ge 2$ then

$$C_G\left(\alpha \right){\cong }^2{A}_m\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{{\displaystyle \frac{sm(m+1)}{2}}}$$

(xi)  

If $G{\cong }^2{D}_m\left(q^{2s}\right)$ for $m\ge 4$ then

$$C_G\left(\alpha \right){\cong }^2{D}_m\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{sm(m-1)}$$

(xii) 

If $G{\cong }^2{E}_6\left(q^{2s}\right)$ then

$$C_G\left(\alpha \right){\cong }^2{E}_6\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{36s}$$

(xiii)

$G{\cong }^3{D}_4\left(q^{2s}\right)$, then

$$C_G\left(\alpha \right){\cong }^3{D}_4\left(r^s\right)$$

So we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=r^{12s}$$

In each subcase, $\left|In{d}_G\left(\alpha \right)\right|$ is divisible by $r^2$.

Case 3: $\alpha$ is a graph automorphism and $r=2$.

By Proposition 4.9.2 in [14], all possibilities for fixed-point subgroups of graph automorphisms of simple groups of Lie type over fields of characteristic 2 are:

(i)  

If $G\cong A_{2m+1}^{\pm }\left(q\right)$ for $m\ge 1$, then $C_G\left(\alpha \right)\cong C_m\left(q\right)$ and we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=q^{m^2+m}$$

(ii) 

If $G\cong A_{2m}^{\pm }\left(q\right)$ for $m\ge 1$, then $C_G\left(\alpha \right)\cong C_m\left(q\right)$ and we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=q^{m^2-m}$$

(iii)

If $G\cong D_m^{\pm }\left(q\right)$ for $m\ge 4$, then $C_G\left(\alpha \right)\cong B_{m-1}\left(q\right)$ and we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=q^{m-1}$$

(iv)

If $G\cong E_6^{\pm }\left(q\right)$, then $C_G\left(\alpha \right)\cong F_4\left(q\right)$ and we have

$${\left|In{d}_G\left(\alpha \right)\right|}_r=q^{12}$$

In each case, $q^2$ is a divisor of $In{d}_G\left(\alpha \right)$.

This leads to a contradiction as $\alpha \in G⟨\alpha ⟩\backslash G$ and $In{d}_G\left(\alpha \right)$ is not square-free.

This final contradiction completes the proof. Therefore, if $In{d}_G\left(x\right)$ is square-free for each $x\in H\backslash G$ where $H=G⟨\alpha ⟩$, then G is solvable. □

4. Examples

In this Section, we give two examples satisfying Theorem 10. It is known that a group is solvable if its order is odd [16] or of the form $p^a{q}^b$ for prime numbers p and q [17]. We consider groups of orders 300 and 1764 in order not to be able to use the above results. To distinguish it from conjugacy class length type of problems, we choose an automorphism $\alpha$ of G which is not an inner automorphism.

Example 2.

Let G be the group $SmallGroup(300,18)$ in the library of [18].

Now, G is a group of order $300=2^2{3}^1{5}^2$. By means of the [18] code at the end of the example, we obtain the following properties of the group G:

$$G=⟨a,b,c,d,e|R⟩$$

where R is the set of relations

$$R=\{d=a^2,ab=ba,ac=ca,ad=da,a{e}^2=ea,b^5=1,$$

$$bc=cb,bd=db,be=eb,c^5=1,cd=dc,ce=ec,$$

$$d^2=1,de=ed,e^3=1\}$$

We choose the noninner automorphism $\alpha :G\to G$ defined on the generators of G by

$$\alpha \left(a\right)=ad$$

$$\alpha \left(b\right)=b^{4}c$$

$$\alpha \left(c\right)=c$$

$$\alpha \left(d\right)=d$$

$$\alpha \left(e\right)=e$$

Now, order of α is 2. We define the semidirect product group $S=G⟨\alpha ⟩$ and consider the set of centralizers of every element in $S\backslash G$. The set of indices of these elements forms the set $indices=\{10,30\}$.

Therefore, for every $x\in G⟨\alpha ⟩\backslash G$ we have either ${Ind}_G\left(x\right)=10$ or ${Ind}_G\left(x\right)=30$ which satisfies the premises of Theorem 10.

When we check the solvability of G by [18], we see G is solvable.

Therefore, the group G and the automorphism α form a nontrivial pair that satisfies the theorem.

We use the following sequence of codes in [18] to find the example explained above:

• G:=SmallGroup(300,18);
• A:=AutomorphismGroup(G);
• H:=Elements(A);
• I:=InnerAutomorphismGroup(G);
• Order(I);
• J:=Difference(A,I);
• Order(J[2760]);
• alpha:=J[2760]^3;
• alpha in I;
• K:=Group(alpha);
• S:=SemidirectProduct(K,G);
• B:=GeneratorsOfGroup(S);
• C:=Group(b[2],b[3],b[4],b[5],b[6]);
• Order(C);
• IsomorphismGroups(C,G);
• U:=Difference(S,C);
• indices:=[];
• for i in U do
• cen:=Centralizer(S,i);
• int:=Intersection(C,cen);
• fix:=Size(int);
• ind:=Order(G)/fix;
• Add(indices,ind);
• od;
• indices;
• IsSolvableGroup(G);

Example 3.

Let G be the group $SmallGroup(1764,23)$ in the library of [18].

Now, G is a group of order $1764=2^2{3}^3{7}^2$. By means of the [18] code at the end of the example, we obtain the following properties of the group G:

$$G=⟨a,b,c,d,e,f|R⟩$$

where R is the set of relations

$$R=\{c=a^2,ab=ba,ac=ca,ad=da,a{e}^2=ea,a{f}^2=fa,d=b^7,$$

$$bc=cb,bd=db,be=eb,bf=fb,c^2=1,cd=dc,ce=ec,$$

$$cf=fc,d^7=1,de=ed,df=fd,e^3=1,ef=fe,f^3=1\}$$

We choose the noninner automorphism $\alpha :G\to G$ defined on the generators of G by

$$\alpha \left(a\right)=ac{e}^2{f}^2$$

$$\alpha \left(b\right)=b$$

$$\alpha \left(c\right)=c$$

$$\alpha \left(d\right)=d$$

$$\alpha \left(e\right)=e^{2}f$$

$$\alpha \left(f\right)=f$$

Now, order of α is 2. We define the semidirect product group $S=G⟨\alpha ⟩$ and consider the set of centralizers of every element in $S\backslash G$. The set of indices of these elements forms the set $indices=\left\{6\right\}$.

Therefore, for every $x\in G⟨\alpha ⟩\backslash G$ we have ${Ind}_G\left(x\right)=6$ which satisfies the premises of Theorem 10.

When we check the solvability of G by [18] we see G is solvable.

Therefore, the group G and the automorphism α form a nontrivial pair that satisfies the theorem.

We use the following sequence of codes in [18] to find the example explained above:

• G:=SmallGroup(1764,23);
• iso:=IsomorphismFpGroup(G);
• F:=Image(iso);
• rels:=RelatorsOfFpGroup(F);
• A:=AutomorphismGroup(G);
• H:=Elements(A);
• I:=InnerAutomorphismGroup(G);
• Order(I);
• J:=Difference(A,I);
• Order(J[457]);
• alpha:=J[457]^21;
• alpha in I;
• K:=Group(alpha);
• S:=SemidirectProduct(K,G);
• B:=GeneratorsOfGroup(S);
• C:=Group(B[2],B[3],B[4],B[5],B[6],B[7]);
• Order(C);
• IsomorphismGroups(C,G);
• U:=Difference(S,C);
• indices:=[];
• for i in U do
• cen:=Centralizer(S,i);
• int:=Intersection(C,cen);
• fix:=Size(int);
• ind:=Order(G)/fix;
• Add(indices,ind);
• od;
• indices;
• IsSolvableGroup(G);

5. Conclusions

As we mentioned in the introduction, the main theorem of this paper is the joining of two types of problems in the area of obtaining information about the structure of a group using arithmetical conditions on the lengths of conjugacy classes and the orders or indices of fixed-point subgroups of automorphisms.

Let G be a group and $\alpha$ be an automorphism of G. In the case of obtaining information on the structure of G using arithmetical conditions on the lengths of conjugacy classes, two important steps are a group with a conjugacy class of length $p^a$ for some prime number p [6] and all conjugacy class sizes are square-free [19]. These results are used to obtain further results. As we mention in the introduction, solvability of $[G,\alpha ]$ is proved when ${Ind}_G\left(\alpha \right)=p^a$ in [6] and when $gcd\left(\right|G|,|\alpha \left|\right)=1$ and ${Ind}_G\left(\alpha \right)$ is square-free in [7]. Theorem 10 proves the latter case by using further restrictions to remove the relatively prime condition. We may consider the statement of Theorem 10 from another point of view as follows:

$H=G⟨\alpha ⟩$ is a group and G is a normal subgroup of H. If the lengths of all conjugacy classes in $H\backslash G$ are square-free, then G is solvable.

This leads us to consider whether we can obtain information about the structure of a normal subgroup by means of arithmetical conditions on the conjugacy classes outside the normal subgroup.

As future work, one may consider the structure of a group H and a normal subgroup G of H when some arithmetical conditions on the lengths of conjugacy classes of $H\backslash G$ are given.