# Is $A_n\times C_2\not\cong S_n$?

I'm working on this exercise that appears in my group theory syllabus:

Prove $A_n\times C_2\not\cong S_n$ for $n\geq3$.

Since the chapter is about normal subgroups and factor groups, I might need to use that $A_n\lhd S_n$, but I don't see how to apply this. I found on the internet several proofs that the semi-direct product between $A_n$ and $C_2$ is isomorphic to $S_n$, but I have not yet seen semi-direct products and automorphism groups, so even if that applies here I'm looking for an answer not using this.

Actually I showed before the exact opposite, but there was a large mistake in my proof.

We assume $n\in\mathbb{Z}$ such that $n\geq3$. Let $G_1=A_n$ and $G_2=\{(1),(1 \ 2)\}$, and then define $H_1=G_1\times\{(1)\}$ and $H_2=\{(1)\}\times G_2$. The following theorem is given and proved in my syllabus:

Let $G$ be a group and $H_1,H_2\subset G$ subgroups such that

a) $h_1h_2=h_2h_1 \ \forall h_1\in H_1,\ \forall h_2\in H_2$;

b) $H_1\cap H_2=\{e\}$;

c) $\forall g\in G, \ \exists h_1\in H_1,h_2\in H_2, \ g=h_1h_2.$

Then $G\cong H_1\times H_2$.

However, clearly a) is not satisfied but I assumed before that it was.

Can anyone give a hint or partial answer? I don't see what to do.

• Think about centres. – Lord Shark the Unknown Mar 11 '18 at 11:45
• So for $n=3$, $A_3\times C_2$ is abelian while $S_3$ is not, and for $n>3$, the centre of both $A_n$ and $S_n$ is trivial (containing only the identity permutation), so $Z(A_n\times C_2)=\{Id_X\}\times C_2$ while $Z(S_n)=\{Id_X\}$. Thus because the centre of $A_n\times C_2$ contains two elements and the centre of $S_n$ only one, both groups cannot be isomorphic. Am I right for these claims? – Václav Mordvinov Mar 11 '18 at 11:53
• Yes, but you really only need to observe that the centre of $S_n$ is trivial and that of $A_n\times C_2$ isn't (for $n\ge3$). – Lord Shark the Unknown Mar 11 '18 at 11:55
• Yes you're right, that makes it even easier! Thanks for this, good hint that didn't spoil too much! :) – Václav Mordvinov Mar 11 '18 at 11:56

There are several possible proofs. The easiest is perhaps to use that $$Z(A_n\times C_2)=Z(A_n)\times Z(C_2)=1\times C_2\cong C_2$$ for all $n\ge 4$, but $Z(S_n)=1$ is trivial for all $n\ge 3$.

References:

The center of $A_n$ is trivial for $n \geq 4$

Find the center of the symmetry group $S_n$.

• Thanks, actually the same as the comment by Lord Shark the Unknown. The only thing that I have to do is prove that $Z(S_N)=\{Id_X\}$ and this is (almost) trivial. – Václav Mordvinov Mar 11 '18 at 12:01
• @VáclavMordvinov No, you also might want to prove that $Z(G\times H)=Z(G)\times Z(H)$. If you think it is trivial that $S_n$ has trivial center, then your homework question above is also trivial. – Dietrich Burde Mar 11 '18 at 12:03
• Using Lord Shark the Unknown's second comment, we see that the center of $A_n$ contains (at least) two elements, so if the center of $S_n$ is indeed trivial, we're done, right? And indeed not trivial but almost imo, trying to prove this right now. Making it formal is not completely trivial indeed! – Václav Mordvinov Mar 11 '18 at 12:05
• No, this is false, see my first link. The center of $A_n$ does not contain at least two elements for $n\ge 4$. But the center of $A_n\times C_2$ does! – Dietrich Burde Mar 11 '18 at 12:06
• Yes I agree, that was a typo actually, sorry! I meant because $Z(C_2)=C_2$, $Z(A_n\times C_2)$ contains at least two elements. – Václav Mordvinov Mar 11 '18 at 12:10