Let $\sigma \in S_n∖A_n$ ($\sigma$ not in $A_n$), prove that the order of $\sigma$ is even.

Question

Given $\sigma \in S_n\setminus A_n$, prove that the order of $\sigma$ is even.

I feel that I have a way to prove it:

Since $\sigma \notin A_n$, the sign of $\sigma$ is $-1$.

This implies that $(-1)^{n-t}=-1$, where $t$ is the number of foreign symmetries that assemble $\sigma$. This gives

$$n-t\equiv 1\bmod 2$$

and then I'm stuck.

I will be glad to see somebody continue or prove it in another way.

Answer 1

$\sigma$ can be written in the products of disjoint $k$-cycles $\sigma_1 \cdots \sigma_r$. Then $|\sigma|=[|\sigma_1|, \cdots, |\sigma_r|]$. Note $|\sigma_i|$ is odd if and only if $\sigma_i \in A_n$. Hence if $|\sigma|$ is odd, then $|\sigma_i|$ are even permutation, and $\sigma \in A_n$. This implies $|\sigma|$ is even.

Answer 2

$A_n$ is normal in $S_n$ and $S_n / A_n$ is the group of order 2. The kernel of the quotient map $\theta : S_n \longrightarrow S_n / A_n$ is $A_n$. So under $\theta$, $S_n \setminus A_n$ maps to the element of order 2 in $S_n / A_n$. Hence every element in $S_n \setminus A_n$ must have even order.