# $\mathbb S_n$ as semidirect product

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.

• Whenever a group $G$ is presented as a semidirect product $H \rtimes K$, where $H$ and $K$ are subgroups of $G$ (or, after the fact, are identified with subgroups of $G$), with $H$ normal, the morphism $\rho \colon K \to \mathrm{Aut}(H)$ is always the same one. It's $K$ acting on $H$ by conjugation. In this case, you just need to find a subgroup $K$ of $S_n$ that is isomorphic to $\mathbf{Z}_2$ and has trivial intersection with $A_n$, and then let $K$ act on $A_n$ by conjugation. – Dave Sep 2 '14 at 5:46
• Well, any subgroup of the form $\{e,(ab)\}$ is isomorphic to $\mathbb Z_2$ and is clearly disjoint from $A_n$. Now, you say to define $\rho(e)=e\tau e$, $\rho((ab))=(ab)\tau(ab)$ for all $\tau \in A_n$. I am not so sure how can I check from here that $\mathbb S_n$ is a semidirect product of $A_n$ by $\mathbb Z_2$. – user16924 Sep 2 '14 at 6:09
• You probably mean $\rho(e)(\tau)$, etc. That's correct. Check that $(h,k) \mapsto hk$ is an isomorphism from $H \rtimes K$ to $G$. (Actually, surjectivity follows from the abstract criteria above only when you assume $G$ to be finite and $|H||K| = |G|$.) – Dave Sep 2 '14 at 6:20

## 3 Answers

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

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

• But the question seems to be mainly about the splitting of that sequence, not the existence. – Tobias Kildetoft Sep 2 '14 at 9:56
• The OP did not seem to know how to start, and the sequence precisely suggests to take a morphism $f : \mathbb{Z}_2 \to S_n$ so that $\mathrm{sign} \circ f = \mathrm{Id}$, and define $\rho : x \mapsto (y \mapsto f(x)yf(x)^{-1})$. But it's my opinion. – Seirios Sep 2 '14 at 13:13
• @TobiasKildetoft Can you give me some details on how to proceed to show that the sequence splits? And if it does split, doesn't that mean that $S_n \cong \mathbb{Z}_2 \oplus A_n$?? And we want to the semi direct product?? Or from that, we get that $\frac{S_n}{A_n} \cong \mathbb{Z}_2$ and so $A_n \triangleleft S_n$ which is really what this was about..? Well actually that would come from being of index 2.... So for $S_n$ to be a semi direct product of the two, it seems the 3 conditions are all pretty obviously fulfilled. Is there something I am missing? – Mutated Penguin Jul 10 '19 at 22:30

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 $A_n$.

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