Semidirect Products of $A_n$ and $\mathbb{Z}/2$ up to isomorphism

Question

Several posts on this site mention that the symmetric group $S_n$ is the semidirect product of $A_n$ and $\mathbb{Z}/2$. Is it true that $S_n$ is the only nontrivial semidirect product $A_n \rtimes \mathbb{Z}/2$ up to isomorphism (perhaps apart from exceptional values of $n$)?

Answer 1

Yes, this is true due to the fact that $\mathrm{Aut}(A_n) = S_n$ via the conjugation action, with exceptions for $n = 1, 2, 3, 6$. An outline of the proof of this fact can be found on Wikipedia.

A nontrivial semidirect product $A_n \rtimes \mathbb{Z}/2$ is determined by a choice of nontrivial homomorphism $\mathbb{Z}/2 \to \mathrm{Aut}(A_n)$; hence we identify $\mathbb{Z}/2$ with a subgroup of $S_n$ of order $2$. Using this identification, the group law on the semidirect product is given by $(a_1, b_1)(a_2, b_2) = (a_1 b_1 a_2 b_1^{-1}, b_1 b_2)$. Hence we obtain an homomorphism $A_n \rtimes \mathbb{Z}/2 \to S_n$ by sending $(a, b) \mapsto ab$, since $$(a_1b_1)(a_2b_2) = a_1b_1a_2b_1^{-1}b_1b_2.$$ If this homomorphism is injective, then it is clearly an isomorphism. Otherwise, it has kernel of order $2$, yielding a normal subgroup of $A_n \rtimes \mathbb{Z}/2$ of order $2$ complimentary to $A_n$. In this case, you can show that the semidirect product is isomorphic to the direct product.

You can check that the result also holds for the case $n = 3$. As the other answer mentions, you do get another semidirect product for $n = 6$. It clearly fails for $n = 1$, since $A_1 = S_1$, and for $n = 2$ the product is direct.

Answer 2

This is true, apart from $n=6$ in which case there is a second, nonisomorphic, such product (and a further extension of order $720$). A proof of this fact is for example in Wilson's GTM text about simple groups.