Show that $S_n \cong A_n \rtimes C_2$

Question

I want to show that $S_n \cong A_n \rtimes C_2$.

Take a transposition $\tau \notin A_n$. Then it is clear that

$$\langle \tau\rangle \cap A_n = 1$$ $$A_n \tau = S_n$$ $$A_n \unlhd S_n$$

and thus $$S_n \cong A_n \rtimes_\phi \langle \tau \rangle$$

with $$\phi: \langle \tau \rangle \to Aut(A_n): \mathcal{\tau} \mapsto (A_n \to A_n: \sigma \mapsto \tau^{-1} \sigma \tau = \tau \sigma \tau), 1 \mapsto 1$$

I'm now trying to see what $\tau^{-1} \sigma \tau$ looks like, as I would like to write:

$$A_n \rtimes_\phi \langle \tau \rangle \cong A_n \rtimes_\psi C_2$$

for some homomorphism $\psi: C_2 \to Aut(A_n)$.

How can I proceed?

Answer 1

Being of index $2$, $A_n$ is normal in $S_n$ and from the inclusion we get the short exact sequence $$\tag1 1\to A_n\to S_n\to C_2\to 1.$$ As you said, we can pick a transposition $\tau\in S_n$, so $\tau^2=1$ and $\tau\notin A_n$. Then sending the nontrivial element of $C_2$ to $\tau$ makes $(1)$ split, which means that $S_n$ is a semi-direct product of $A_n$ by $C_2$.