# Show that, for every $n$, $A_{n+2}$ has a subgroup isomorphic to $S_n$

Show that, for every $n$, $A_{n+2}$ has a subgroup isomorphic to $S_n$

Also, in general, when I construct an isomorphism, what's necessary to show that it's well-defined? Is showing it is a bijection and homomorphism enough? And are there any rules to follow when I try to construct a homomophism or isomorphism? I mean when I'm asked to show a certain group is isomorphic to another, it's always difficult for me to find the mapping.

Here is a more detailed explanation of the construction suggested in Tobias Kildetoft's answer. Consider $S_n$ as a subgroup of $S_{n+2}$ in the obvious way (namely, as the permutations of $\{1,\dots,n+2\}$ that fix $n+1$ and $n+2$), and let $t\in S_{n+2}$ be the transposition that swaps $n+1$ and $n+2$. Note that $t$ commutes with every element of $S_n\subset S_{n+2}$. Define a map $\varphi:S_n\to S_{n+2}$ by $\varphi(x)=x$ if $x$ is an even permutation and $\varphi(x)=xt$ if $x$ is an odd permutation. Then $\varphi$ is a homomorphism:

• If $x$ and $y$ are even, then $xy$ is even, and $\varphi(xy)=xy=\varphi(x)\varphi(y)$.
• If $x$ is odd and $y$ is even, then $xy$ is odd and $\varphi(xy)=xyt=xty=\varphi(x)\varphi(y)$ (here we use that $t$ commutes with every element of $S_n$).
• If $x$ is even and $y$ is odd, then $xy$ is odd, and $\varphi(xy)=xyt=\varphi(x)\varphi(y)$.
• If $x$ and $y$ are odd, then $xy$ is even, and $\varphi(xy)=xy=xyt^2=xtyt=\varphi(x)\varphi(y)$ (here we use that $t^2=1$ and $t$ commutes with every element of $S_n$).

It is also clear that $\varphi$ is injective, because $\varphi(x)$ maps $\{1,\dots,n\}$ to itself for any $x\in S_n$ and we can recover $x$ as the restriction of $\varphi(x)$ to $\{1,\dots,n\}$. Finally, the image of $\varphi$ is contained in $A_{n+2}$, since if $x$ is even, then $\varphi(x)=x$ is even, and if $x$ is odd, then $\varphi(x)=xt$ is even since $t$ is odd.

Thus $\varphi$ is an isomorphism from $S_n$ to the subgroup $\varphi(S_n)\subseteq A_{n+2}$.

Hint: The even permutations in $S_n$ are already there as the ones that fix two given elements. Now try to add to these the same ones, but which instead of fixing those two elements, transposes them. Show that you now have the desired subgroup.

Edit: As pointed out by Anon, I messed up the hint. For the second part, one should take an odd permutation and compose it with the transposition of those two given elements.

• Thank you, I messed up that part. Edited. – Tobias Kildetoft Dec 30 '12 at 3:17
• +1. To elaborate symbolically, fix two elements $a,b\in X$ with $|X|=n+2$ and write $Y=X\setminus\{a,b\}$. Then $\mathrm{Sym}(X)\cong S_{n+2}$ and $\mathrm{Sym}(Y)\cong S_n$; view $A_n$ as a subgroup of $S_n$ as a subgroup of $S_{n+2}$. Then consider $A_n\cup (ab)(S_n\setminus A_n)$ - this is a subgroup of even permutations isomorphic to $S_n$ and so is contained in $A_{n+2}$. – anon Dec 30 '12 at 3:24
• @anon You're right about my statements (in my deleted proof). For some reason, that was the motivation to me, but it doesn't work out. – Calvin Lin Dec 30 '12 at 7:24
• @CalvinLin If you mod out $S_n$ by the subgroup generated by an odd permutation of order two, then there is indeed a transversal which is a subgroup, since then each coset contains an even and an odd permutation and the numbers match up. This seems to be a bit of a coincidence, though. – Tobias Kildetoft Dec 31 '12 at 1:12