$\mathbb S_n$ as semidirect product

Question

In this note, I've read that $\mathbb S_n$ is a semidirect product of the alternating group $A_n$ by $\mathbb Z_2$. So I am trying to define a morphism $\rho: \mathbb Z_2 \to Aut(A_n)$ to show that $\mathbb S_n \cong A_n \rtimes Z_2$. I would appreciate suggestions on how could I define the morphism. Thanks in advance.

Answer 1

Hint: Given $\mathrm{sign}$ the parity of a permutation, you have an exact sequence

$$1 \to A_n \to S_n \overset{\mathrm{sign}}{\longrightarrow} \mathbb{Z}_2 \to 0.$$

Answer 2

What are the permutations in $S_n$ that are not in $A_n$? How do they operate on $A_n$. Take $S_3$ as an example.

Answer 3

More generally:

Theorem. If $G$ has a subgroup of index two, $H$ say, and there exists an element $g\not\in H$ of order two then $G$ splits as a semidirect product $G=H\rtimes\mathbb{Z}_2$.

This is because $\langle g\rangle\cap H=1$, and the other (internal) semidirect product conditions follow because $H$ has index two (so is normal, and so on).

So, in order to answer your question you simply need to find an element of order two in $S_n$ which is not contained in $A_n$.