Is it true that $A_n$ contains all the elements of odd order?
I think yes, but I would like to double check my answer, and ask if there are any alternative proofs.
Take $\sigma \in S_n$ with $|\sigma|$ odd. Now $\sigma$ has a cycle decomposition $\sigma = \sigma_1 … \sigma_m$ into disjoint cycles. Now $|\sigma|= \text{lcm}(|\sigma_1|, …, |\sigma_m|)$. Thus each $|\sigma_i|$ divides $|\sigma|$, so $|\sigma_i|$ is odd, and being a cycle, $\sigma_i$ is in $A_n$. Therefore $\sigma \in A_n$.
Best Answer
An alternative proof uses the fact that $A_n$ is a normal subgroup of index $2$ in the symmetric group $S_n$. So the quotient $S_n/A_n$ is a cyclic group of order $2$. Now if $\sigma\in S_n$ has odd order $k$, and if I write $[\sigma]$ for the equivalence class of $\sigma$ in $S_n/A_n$, then $[\sigma]^k=[\sigma^k]=[e]$, but also, since both elements of $S_n/A_n$ satisfy $x^2=[e]$, and since $k$ is odd, we have $[\sigma]^k=[\sigma]$. So $[\sigma]=[e]$, which means that $\sigma\in A_n$.