# Is it true that $A_n$ contains all the elements of odd order?

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 $$\sigma \in A_n$$.

• I corrected a few simple typos in an otherwise nice proof. – darij grinberg Dec 6 '19 at 2:47
• @darijgrinberg Thanks! – 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 $$\sigma\in A_n$$.
• 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 $\operatorname{sgn}(\sigma) = 1$. – Daniel Schepler Dec 6 '19 at 2:13
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 $$sgn(\sigma)=1$$.