# $A_n \times \mathbb{Z} /2 \mathbb{Z} \ \ncong \ S_n$ for $n \geq 3$

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.

• If you have a bijection $f:S_n\rightarrow A_n\times \Bbb Z/2\Bbb Z$, then $f^{-1}\left((A_n,1)\right)$ is a subgroup of $S_n$, with index 2, so it's $A_n$. – Jean-Claude Arbaut Aug 3 '14 at 20:48
• Commenting on the second part. Your suggestion doesn't work because on the left $(\sigma,1)$ and $(\sigma,-1)$ commute for all $\sigma\in A_n$, so their homomorphic images should also commute. This forces $\phi(1,-1)$ to be the identity (if $n>3$). Also there are automorphisms of $A_n$. Those are relatively tricky to list even though the answer is simple except when $n=6$. As this sounds like an exercise from a first or a second course on abstract algebra, it is fairly sure that you are not expected to go in that direction :-/ – Jyrki Lahtonen Aug 4 '14 at 6:55

In case $S_n=\mathbb{Z}/2\mathbb{Z} \times A_n$, we have $S_n=NA_n$, with $\mathbb{Z}/2\mathbb{Z} \cong N \lhd S_n$, $N \cap A_n=1$. Since $S_{n}'=[S_n,S_n]=A_n$ and $N \cap A_n=1$, we have $N \subseteq Z(S_n)=\{(1)\}$, a contradiction (in general, if $N \lhd G$ and $N \cap G'=1$, then $N \subseteq Z(G)$).
Yet another way of obtaining a contradiction: take the centers at both sides and use that $Z(A_3)=A_3$ and $Z(A_n)=\{(1)\}$ for $n \gt 3$: $$Z(A_n \times \mathbb{Z}/2\mathbb{Z})=Z(A_n) \times \mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z},$$ while $Z(S_n)=\{(1)\}$.
• where did you use $n\ge 3$ though? – Bombyx mori Aug 3 '14 at 21:07
• $Z(S_n)=\{(1)\}$ for $n \geq 3$ – Nicky Hekster Aug 3 '14 at 21:08
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.