# 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. Mar 11, 2018 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? Mar 11, 2018 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 $n\ge3$). Mar 11, 2018 at 11:55
• Yes you're right, that makes it even easier! Thanks for this, good hint that didn't spoil too much! :) Mar 11, 2018 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. Mar 11, 2018 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. Mar 11, 2018 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! Mar 11, 2018 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! Mar 11, 2018 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. Mar 11, 2018 at 12:10