Show $A_n$ is invariant under conjugation
$$\pi \in S_n , \pi A_n \pi^{-1} =A_n $$
$\Rightarrow$] ($\pi A_n \pi^{-1} \subset A_n$)
$x \in S_n $, $x\in \pi A_n \pi^{-1}$ so $\pi x \pi^{-1} \in A_n$ conjugation is a famous automophism $x \in A_n$
another argument would be $\pi x \pi^{-1}$ is even set it equal to $\sigma_{even}$ so $x = \pi^{-1} \sigma \pi $
we go into cases if $\pi$ is even or odd, either case $x$ is even
$\Leftarrow $ similar argument
Studying for a test appreciate feedback, thanks
I genuinely don't understand anything you've written down. How about if we try this:
Since $S_n$ is generated by transpositions, it is enough to show that $\tau A_n\tau=A_n$ when $\tau=\tau^{-1}$ is a transposition. Now, if $\sigma\in A_n$, then $\sigma$ can be written as a product of an even number of transpositions, say $$\sigma=\tau_1\cdots\tau_{2k}.$$ Now, $\tau\sigma\tau=\tau(\tau_1\cdots\tau_{2k})\tau$ is a product of $2k+2$ transpositions, hence belongs to $A_n$.
This shows that $\tau A_n\tau\subset A_n$. But, both sets have order $n!/2$, so they are equal.