TITLE: $A_n \times \mathbb{Z} /2 \mathbb{Z} \ \ncong \ S_n$ for $n \geq 3$
QUESTION [9 upvotes]: I try to show that $A_n \times \mathbb{Z} /2 \mathbb{Z} \  \ncong \ S_n$ for $n \geq 3$.

It is not hard to show the statement for $n=3$.
We have 
$$
A_3 \times \mathbb{Z} /2 \mathbb{Z} \ \cong \  
\mathbb{Z} /3 \mathbb{Z} \times \mathbb{Z} /2 \mathbb{Z} \ \cong \ 
\mathbb{Z} /6 \mathbb{Z} 
$$
an abalian group, while $(123)(12) = (13) \neq (23)= (12)(123)$ shows that $S_3$ is not abelian.

Here I try to solve the exercise in general. I guess that the only homomorphism
$$
\phi \quad : \quad A_n \times \mathbb{Z}/2\mathbb{Z} \longrightarrow S_n 
$$
can be made my mapping $(\sigma,1) \mapsto \sigma$ and $(\sigma,-1) \mapsto (\tau \sigma)$ for some odd permutation $\tau$. I don't know how I could prove that. Please give me a hint to go on.

REPLY [2 votes]: Let $\alpha\in S_n$ with $\alpha\not = 1$. Choose some $i = 1, \dots, n$ such that $j = \alpha(i)$ has $j\not = i$. Set $\beta = (jk)$ for $k\not = i, j$. Then $\alpha(i) = j$ but $\beta^{-1}\alpha\beta(i) = \beta^{-1}(j) = k$. Hence $\beta^{-1}\alpha\beta\not= \alpha$, and $\alpha$ is not central. Since $\alpha$ is arbitrary, we have $Z(S_n) = 1$, from which your statement follows.