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.

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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.
+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
@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