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 $\sigma \in A_n$.

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    $\begingroup$ I corrected a few simple typos in an otherwise nice proof. $\endgroup$ – darij grinberg Dec 6 '19 at 2:47
  • $\begingroup$ @darijgrinberg Thanks! $\endgroup$ – Ovi Dec 6 '19 at 3:43

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 $\sigma\in A_n$.

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    $\begingroup$ Or to say essentially the same thing, if $\sigma^n = e$ with $n$ odd, then $\operatorname{sgn}(\sigma)^n = 1$ and $\operatorname{sgn}(\sigma) \in \{ \pm 1 \}$ implies $\operatorname{sgn}(\sigma) = 1$. $\endgroup$ – Daniel Schepler Dec 6 '19 at 2:13

That is exactly correct and I don't think there is a more clear answer.

  • $\begingroup$ +1 Thanks! I'd also be interested in less clear answers, just to become more familiar with group theory. $\endgroup$ – Ovi Dec 6 '19 at 2:07

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 $sgn(\sigma)=1$.


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