Subgroup of $S_{n}$ that is not in $A_{n}$ I'm trying to prove this following statement:
Let $G\leq S_{n}$ for $n\geq 5$.
I need to show that if $G\not\subseteq A_{n}$ then $GA_{n}=S_{n}$.
I was thinking to show it implies $G\cap A_{n}=e$, and then using direct product, but I don't know if it is necessarily correct. If it doesn't, can I get a hint?
Thank you.
 A: As $G \not \subseteq A_n $, there exists an odd permutation $\sigma \in G $. We can see that $\sigma A_n \subseteq GA_n$ must then consist of only odd permutations, and since it has $\frac{n!}{2}$ elements, it consists of all odd permutations. Moreover, $e \in G$, and thus, $eA_n \subseteq GA_n$ consists of all even permutations. Hence, $GA_n=S_n$.
A: Observe that, since $A_n$ is normal, $GA_n$ is necessarily a subgroup of $S_n.$
Also $|G|\ge2$ and $|G\cap A_n|=1$ gives us $$|GA_n|=\dfrac{|G||A_n|}{|G\cap A_n|}=?$$
In fact, this tells you more about the order of such groups $G.$
A: We have $[S_n:A_n]=2$, so $GA_n$ can only be $A_n$ or $S_n$, since there are no other subgroups in between $A_n$ and $S_n$. Since $G\not\subseteq A_n$, we cannot have $GA_n=A_n$, so it must be $S_n$.
A: The hypothesis $n \geqslant 5$ is not necessary. For $n \geqslant 2$ we have that $(\Sigma_n\colon\mathrm{A}_n)=2$ and we appeal to the following general:

Lemma. Let $G$ be an arbitrary group and $H \leqslant G$ be a subgroup such that the index $(G \colon H)$ be a prime natural number. Then $H$ is a maximal subgroup of $G$ (i.e. a maximal element of the set of proper subgroups of $G$ ordered by inclusion).

We gather in particular that the alternate group $\mathrm{A}_n$ is maximal in $\Sigma_n$. If $\Gamma \leqslant \Sigma_n$ we have on the one hand that $\Gamma\mathrm{A}_n \leqslant \Sigma_n$ by normality of $\mathrm{A}_n$. By virtue of the maximality of $\mathrm{A}_n$, this relation entails $\Gamma\mathrm{A}_n=\Sigma_n$ or $\Gamma\mathrm{A}_n=\mathrm{A}_n$, the latter being equivalent to $\Gamma \leqslant \mathrm{A}_n$.
For $n=0, 1$ the statement is trivially true, as $\Sigma_0$ and $\Sigma_1$ are both trivial groups. The claim you make generalises to finitary groups of permutations $\Sigma_0(A)$ on arbitrary sets $A$.
