TITLE: Is it true that $A_n$ contains all the elements of odd order?
QUESTION [9 upvotes]: 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 $\sigma \in A_n$.

REPLY [2 votes]: Your proof is correct; another approach is as follows.
Restrict the sign homomorphism to the subgroup generated by given permutation $\sigma$.
$$\phi:\langle\sigma\rangle\to\{\pm1\}$$
Order of codomain is $2$, order of domain is odd, and we know that order of image divides both of them. Therefore, $\phi$ is trivial homomorphism and hence $sgn(\sigma)=1$.