Normal subgroups of $S_n$ for $n\geq 5$.

Question is to :

Find all normal subgroups of $S_n$ for $n\geq 5$.

What I have done so far is :

We know that $A_n$ is one normal subgroup of $S_n$.

Suppose $H\neq (1)$ is another normal subgroup of $S_n$ then,

$H\cap A_n$ would also be a normal subgroup.

As $H\cap A_n \subseteq A_n$ we see that $H\cap A_n \unlhd A_n$.

But for $n\geq 5$, $A_n$ is a simple group so does not have proper normal subgroups.

Thus, $H\cap A_n =(1)$. So, we should have $A_n\leq H \leq S_n$ with $|A_n|=\frac{n!}{2}$ and $|S_n|=n!$

As there is no number in between $\frac{n!}{2}$ and $n!$ which is divisible by $\frac{n!}{2}$ and divides $n!$ we should end up with the case when $H=A_n$ or $H=S_n$.

I feel thankful and it would be appreciated if someone can proof read this solution

P.S : I have checked for this problem in this form and got two questions (one is tagged as duplicate) the original question do not have any proof for the Question "Find all normal subgroups of $S_n$ for $n\geq 5$". So, please do not tag it as duplicate as the question is not answered in any of the other OP.

Solution 1:

Your step "$H \cap A_n = 1$ implies $A_n \leq H \leq S_n$" is too fast: you have to rule out the case $H \cap A_n = 1$. This can be done as follows: in general if $H \unlhd G$ with $H \cap G' = 1$, then $H \subset Z(G)$. Since $Z(S_n) = 1$ for $n \geq 3$, you can conclude $H=1$ in your case.
Further, in general if $K$ is a subgroup of $G$ with $index[G:K]=2$, then from $K \leq H\leq G$ it follows that $K=H$ or $H=G$. So the rest of your proof is correct!