Show a group is a semi direct product

[DeldotB]

Homework Statement

Good day,

I need to show that $S_n=\mathbb{Z}_2$(semi direct product)$Alt(n)$
Where $S_n$ is the symmetric group and $Alt(n)$ is the alternating group (group of even permutations) note: I do not know the latex code for semi direct product

Homework Equations

none

The Attempt at a Solution

(I think) It suffices to show that $\mathbb{Z}_2\cap Alt(n)=0$ (where 0 is the identity) and that
$S_n= \mathbb{Z}_2 Alt(n)$. My question is, is $S_n= \mathbb{Z}_2 Alt(n)$ even valid notation? And how do I begin to do this? p.s I get this notation from my book which says:

To show a group G is a semi direct product, show $G=NH$ and $N \cap H$= identity.

I should mention here that the alternating group is the normal subgroup (I think).

For the first part, since $\mathbb{Z}_2$= {0,1}, its pretty clear that its intersection with Alt(n) is just the identity. I am having problems with the second part..

 

[andrewkirk]

The latex code for semidirect product is \rtimes.

The problem's notation is muddled. It should ask you to prove either

(1) isomorphism, not equality, ie to prove that $S_n\cong \mathbb{Z}_2\rtimes Alt(n)$, where the semidirect product is Outer.
OR
(2) equality, where the semidirect product is Inner, ie $S_n=N\rtimes Alt(n)$ where $N$ is any subgroup of $S_n$ of order 2, which hence must be isomorphic to $\mathbb{Z}_2$.

I suggest trying for the second one.

It suffices to show that $\mathbb{Z}_2\cap Alt(n)=0$ (where 0 is the identity) and that
$S_n= \mathbb{Z}_2 Alt(n)$. My question is, is $S_n= \mathbb{Z}_2 Alt(n)$ even valid notation?

The notation is valid. If $G,H$ are subgroups $GH$ is defined as the set of all elements that can be written as $gh$ where $g\in G,\ h\in H$. That is not necessarily a subgroup. So part of what you have to show is that $S_n= \mathbb{Z}_2 Alt(n)$ (or rather $N\,Alt(n)$ using my notation of (2) above) is a subgroup.

Why not pick N to be the subgroup generated by the swap permutation (1 2). Then try to prove the two things you need to prove. The intersection one is dead easy. The other, not so much.