TITLE: Subgroup of $A_{n}$
QUESTION [5 upvotes]: Suppose $G \leq S_{n}$ and $G$ has an odd number of elements. Prove $G \leq A_{n}$. I'm trying to do this by contradiction by assuming $G$ has an odd permutation but I can't show how it would be in $A_{n}$.

REPLY [4 votes]: This has nothing to do with odd permutations. Instead, you may consider the quotient map $p : S_n\rightarrow S_n/A_n = C_2$. Now ask yourself, what is the image of $G$ under this surjection? If $G$ surjects onto $C_2$, then it must have even order, thus $p(G) = 1$ in $C_2$, implying that $G$ is contained in the kernel of $p$, hence $G\le A_n$.