2 problems about the symmetric group 
*

*Let $H$ be a subgroup of $S_{n}$ where $n\geq 2$. Then, every element of $H$ is even, or exactly half of the elements of $H$ are even.

*Prove that $A_{n}$ contains a subgroup isomorphic to $S_{n-2}$ for each $n$, where $n\geq 3$.
Can you give me a hint please? I don't know how to start the proof.
 A: 1) Consider the group homomorphism $f:H\to \{\pm 1\}$ mapping $h\in H$ to $1$ if $h$ is even, and to $-1$ if $h$ is odd. Its kernel is the set of even elements in $H$. If not all elements in $H$ are even, then $f$ is epi (i.e., surjective). Now use the first isomorphism theorem and deduce what the size of the kernel is. 
2) Find some inspiration by identifying $S_2$ in $A_4$, and $S_3$ in $A_5$. Learn to get your hands dirty this way. 
A: *

*Partition $H$ into its even and odd permutations. These will be the cosets of an index two subgroup of $H$ (consider its intersection with $A_n$).

*View $S_{n-2}$ as a subgroup of $S_n$. Perhaps we can keep the even elements of $S_{n-2}$ but change the odd elements to even ones and still end up with an isomorphic subgroup; what about multiplying the odd elements by a fixed odd permutation so that they all become even? It'd make things easier if this odd permutation $\sigma$ didn't "interact" with anything in $S_{n-2}$ badly; what if we let it only permute things outside $\{1,2,\cdots,n-2\}$? There happens to be only one odd permutation that can do that (it's a certain transposition). Check that this works.

