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By a $\mathbb{Z}_2$ extension of $A_n$, I mean the following short exact sequence:

$1 \longrightarrow A_n \longrightarrow G \longrightarrow \mathbb{Z}_2 \longrightarrow 1$.

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

Note: One can check that if the above S.E.S. splits then there are only two choices of $G= S_n$ or $ A_n\times\mathbb{Z}_2$, up to isomorphism, since $\mathrm{Out}(A_n)=\mathbb{Z}_2$. But I'm not able to show that the above S.E.S. always splits i.e. there is no non-split group extension of $A_n$ by $\mathbb{Z}_2$.

Rajesh Dey
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  • Your terminology is backwards; this should be called an $A_n$ extension of $\mathbb{Z}_2$. – Qiaochu Yuan Oct 01 '23 at 21:13
  • @QiaochuYuan Not necessarily. There is a split among group theorists on whether a group $G$ with normal subgroup $N$ and $G/N\cong K$ is "an extension of $N$ by $K$", or "an extension of $K$ by $N$". – Arturo Magidin Oct 02 '23 at 00:32
  • One way out of this terminological conundrum is to drop the "e" word, read the short exact sequence from left to right, and say that $G$ is an "$A_n$-by-$\mathbb Z_2$ group". – Lee Mosher Oct 02 '23 at 11:24

1 Answers1

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Let $G$ be such an extension. If the centralizer of $A_n$ in $G$ is nontrivial then, since $A_n$ has trivial centre, this centralizer must have order $2$ and it would be a normal subgroup giving $G \cong A_n \times C_2$.

Otherwise the centralizer is trivial, in which case conjugation induces an injective homomorphism $G \to S_n$, and so $G \cong S_n$, which is a split extension.

Added later: Note that the above is true only when $n \ne 6$. When $n=6$ there are four extensions of this type, three of which are split and one nonsplit.

Rajesh Dey
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Derek Holt
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  • It also follows from one of your earlier answers https://math.stackexchange.com/a/3658047/1226122 :-The extensions are in one-one correspondence to the number of homomorphisms from $\mathbb{Z}_2\longrightarrow \mathrm{Out}(A_n)$, since the centre of $A_n$ is trivial. Since there are exactly two homomorphisms from $\mathbb{Z}_2\longrightarrow \mathbb{Z_2}$ viz. trivial and id, there are only two extensions. Hence the above S.E.S always splits. Am I right? – Rajesh Dey Oct 01 '23 at 15:44
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    Yes that's right at least when $n \ne 6$. When $n=6$, ${\rm Out}(A_6) \cong C_2 \times C_2$, and there are four such extensions. – Derek Holt Oct 01 '23 at 18:14