Subgroup of $A_{n}$

Question

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 $A_{n}$.

Answer 1

Suppose $G\nleq A_n, \exists g_0 \in G\setminus A_n$. Now consider $$ A = \{g \in G : \text{sgn}(g) = 1\}, \text{ and } B = \{g \in G : \text{sgn}(g) = -1\} $$ Then $A = g_0B$, so $|A| = |B|$. Also, $G = A\sqcup B$, so $|G|$ would be even. This is a contradiction.

Answer 2

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 $G\le A_n$.