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.