Conjugacy-separable with only finitely many prime divisors of orders of elements implies every extensible automorphism is class-preserving

Statement

Suppose $G$ is a Conjugacy-separable group (?): in other words, given any two elements $x,y$ of $G$ that are not conjugate, there exists a normal subgroup of finite index $N$ in $G$ such that the images of $x,y$ in $G/N$ are not conjugate in $G/N$.

Suppose, further, that the set of primes $p$ that divide the order of some non-identity element of $G$ is finite.

Then, if $\sigma$ is an Extensible automorphism (?) of $G$, $\sigma$ is a Class-preserving automorphism (?) of $G$.

Related facts

• Finite-extensible implies class-preserving
• Conjugacy-separable implies every quotient-pullbackable automorphism is class-preserving
• Conjugacy-separable implies every semidirectly extensible automorphism is class-preserving
• Conjugacy-separable and aperiodic implies every extensible automorphism is class-preserving

Facts used

1. Semidirectly extensible implies linearly pushforwardable for representation over prime field
2. Every finite group admits infinitely many sufficiently large prime fields
3. Sufficiently large implies splitting, Splitting implies character-separating, Character-separating implies class-separating

Proof

Given: A conjugacy-separable group $G$ with only finitely many primes $p$ dividing the orders of elements of $G$. An extensible automorphism $\sigma$ of $G$. Two elements $x,y$ of $G$ that are not conjugate.

To prove: $\sigma$ cannot send $x$ to $y$.

Proof:

1. There exists a normal subgroup $N$ of finite index in $G$ such that the images of $x$ and $y$ are not conjugate in $N$: This follows from the definition of conjugacy-separable.
2. Let $\overline{G}=G/N$. Then, there exists a prime $p$ such that the field of $p$ elements is sufficiently large for $\overline{G}$, and such that $p$ does not divide the order of any element of $G$: By fact (2), there are infinitely many sufficiently large prime fields for $\overline{G}$, i.e., there are infinitely many primes $p$ for which the corresponding prime field is sufficiently large for $G$. Since there are only finitely many prime divisors of orders of elements, we can find a prime $p$ not among any of these divisors such that the corresponding prime field is sufficiently large.
3. The field of $p$ elements is a class-separating field for $\overline{G}$. In particular, there is a finite-dimensional linear representation ${\rho }_1:\overline{G}\to GL(V)$ of $\overline{G}$ over this field such that ${\rho }_1(\overline{x})$ and ${\rho }_1(\overline{y})$ are not conjugate: This follows from fact (3).
4. $\sigma$ is linearly pushforwardable over the prime field with $p$ elements, for the $p$ chosen above. In particular, if $\sigma (x)=y$, then $\rho (x)$ and $\rho (y)$ are conjugate for any representation $\rho$ over this field: Let $\rho$ be a representation of $G$ over this field. Let $V$ be the corresponding vector space and $H=V⋊G$ the semidirect product for the action. Since $p$ does not divide the order of any element of $G$, $V$ is the set of elements of $H$ of order dividing $p$. In particular, $V$ is characteristic in $H$, and thus, if $\sigma$ extends to an automorphism ${\sigma }^{\prime }$ of $H$, then ${\sigma }^{\prime }$ also restricts to an automorphism $\alpha$ of $V$. Fact (1) thus yields that $\rho \circ \sigma =c_{\alpha }\circ \rho$, so $\sigma$ is linearly pushforwardable over the field of $p$ elements. In particular, if $\sigma (x)=y$, then $\rho (\sigma (x))=c_{\alpha }(\rho (x))$, so $\rho (y)$ is conjugate to $\rho (x)$ by $\alpha$.
5. Let ${\rho }_1$ be the linear representation chosen in step (3), and let $\rho ={\rho }_1\circ p$ where $p:G\to G/N=\overline{G}$ is the quotient map. Then, $\rho (x)$ and $\rho (y)$ are not conjugate in the general linear group $GL(V)$. However, by step (4), we have that $\sigma$ is linearly pushforwardable, so if $\sigma (x)=y$, then $\rho (x)$ and $\rho (y)$ are conjugate. This gives a contradiction, so we cannot have $\sigma (x)=y$, and we are done.