Extensions of alternating group by cyclic group

Question

For integers $n\geq5$ and $n\neq6$, and $m\geq2$, by an extension $G$ of alternating group $A_n$ by cyclic group $\mathbb{Z}_m$, I mean the following short exact sequence:

$1 \to A_n \to G \to \mathbb{Z}_m \to 1$.

Question: How many $G$ are there up to isomorphism?

Notes:

(1) For $m=2$, it follows from Derek Holt's answer that $G \cong S_n$ or $ A_n\times\mathbb{Z}_2$.

(2) For $m>2$, note that, there are two homomorphisms $\phi:\mathbb{Z}_m \to \mathbb{Z}_2$, namely, $\bar{1} \mapsto \bar{0}$ or $\bar{1} \mapsto \bar{1}$ if $m$ is even, and one homomorphism, namely, $\bar{1} \mapsto \bar{0}$ if $m$ is odd. Since $\mathrm{Out}(A_n)\cong \mathbb{Z}_2$ for $n\neq6$ and $Z(A_n)=1$, therefore every $\phi:\mathbb{Z}_m \to \mathrm{Out}(A_n)$ determine an unique extension and vice versa. For details, see another answers this and this.

From the above answers, can we conclude the following?

Case 1: $m$ is odd. $G \cong A_n\times\mathbb{Z}_m$.

Case 2: $m$ is even. $G \cong A_n\times\mathbb{Z}_m$ or $A_n \rtimes _{\bar{1} \mapsto \bar{1}}\mathbb{Z}_m$.

In particular, for the case $n=5$ and $m=3$, consider the homomorphism $\phi:\mathbb{Z_3}\to \mathrm{Aut}(A_5)$ determined by $\bar{1}\to (1~2~3)$. I'm not able to show that $ A_5\rtimes_{\phi}\mathbb{Z}_3 \cong A_5\times\mathbb{Z}_3$, which would otherwise violate the above conclusion.

Answer 1

The posts you link to give an explicit way of constructing the isomorphism (for instance, ashpool's answer here).

Take $A_5\rtimes C_3$, with the generator $x$ of $C_3$ acting like conjugation by $(123)$, ${}^x\!\sigma= (123)\sigma(123)^{-1}$. Then define $f\colon A_5\rtimes C_5 \to A_5\times C_3$ by $f(\sigma,x^i) = (\sigma(123)^i,x^i)$.

This is well-defined as a function, since $(123)$ and $x$ have the same order. It is a group homomorphism: $$\begin{align*} f\Bigl( (\sigma,x^i)(\tau,x^j)\Bigr) &= f\Bigl( \sigma ({}^{x^i}\!\tau),x^{i+j}\Bigr)\\ &= f\Bigl (\sigma (123)^i\tau (123)^{-i},x^{i+j}\Bigr)\\ &= \Bigl( \sigma(123)^i\tau(123)^{-i}(123)^{i+j},x^{i+j}\Bigr)\\ &= \Bigl( \sigma(123)^i\tau(123)^{j},x^{i+j}\Bigr);\\ f(\sigma,x_i)f(\tau,x^j)&= (\sigma(123)^i,x^i)(\tau(123)^j,x^j)\\ &= (\sigma(123)^i\tau(123)^j,x^{i+j}). \end{align*}$$ If $f(\sigma,x^i) = (e,e)$, then $i=0$, and $\sigma$ is the identity, so the map is one-to-one, hence a bijection.

Thus, $A_5\rtimes C_3\cong A_5\times C_3$, as desired.

The arbitrary case with $n\gt 3$ (so $A_n$ has trivial center), $n\neq 6$ works the same way. If $\phi\colon C_m\to\mathrm{Inn}(A_n)$ maps the generator to conjugation-by-$\rho$, (where $\rho$ has order dividing $m$), you define $f\colon A_n\rtimes C_m\to A_n\times C_m$ by $f(\sigma,x^i) = (\sigma\rho^i,x^i)$. A calculation as above shows that you have an isomorphism.