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 $G$.

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 $H = V \rtimes G$ 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 $\sigma'$ of $H$, then $\sigma'$ 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.