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

Statement
Suppose $$G$$ is a fact about::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 fact about::extensible automorphism of $$G$$, $$\sigma$$ is a fact about::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) uses::Semidirectly extensible implies linearly pushforwardable for representation over prime field
 * 2) uses::Every finite group admits infinitely many sufficiently large prime fields
 * 3) uses::Sufficiently large implies splitting, uses::Splitting implies character-separating, uses::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 $$H = V \rtimes G$$ 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.