A similar question is asked by some user but unanswerd (Semidirect products of $S_n$). I wants to show that the semidirect product of $A_n$ and $\mathbb{Z}_2$ is the symmetric group $S_n$.
$S_n\cong A_n \rtimes \mathbb{Z}_2?$
I tried a lot but I have seen that $A_5\rtimes \mathbb{Z}_2\cong $SL$(2,5)$. Please tell whethere I am correct or not?
$A_n$ is a normal subgroup of $S_n$ (one way to see this is to use that $[S_n : A_n] = 2$, and index $2$ subgroups are always normal). Now take $\sigma$ to be any $2$-cycle, say $\sigma = (12)$. Then since $\sigma$ is an odd permutation, it does not lie in $A_n$; this means the subgroup generated by both $A_n$ and $\langle \sigma\rangle$ must be all of $S_n$ (otherwise, note that its index must be smaller than $2$ but larger than $1$, which is impossible). Finally, note that $A_n\cap \langle \sigma\rangle = \{1\}$.
The above shows that $S_n$ satisfies the "inner semidirect product" conditions, and we therefore have $S_n \cong A_n\rtimes \langle \sigma\rangle$. Now note that $\langle \sigma\rangle \cong \mathbb{Z}_2$.