Solution check request for showing that the alternating group $A_5$ has no subgroup of order 30. Can someone check my solution to the following question please:
Prove that the alternating group $A_5$ has no subgroup of order 30.
I will be assuming the following result:

Theorem 1:  Let $N$ be a subgroup of a group $G$ of index 2.  Prove that $N$ is a normal subgroup.

Solution:  Suppose $A_5$ has a subgroup group $N$ of order 30.  The order of $A_5$ is 60 and by lagrange theorem, $|A_5|/|N|=2$.  Hence by the Theorem 1,  $N$ is of index 2, and so $N$ is normal.  But $A_5$ is a simple group.  Hence we arrive at a contradiction.
I try to do a simple proof where I don't have to look into the cycle structures of $A_5$ like the first question from here: question 1 and here:  question 1, nor making use of the class equation, which Hungerford's Abstract algebra 3rd edition has not introduced at that edition of the text.
Thank you in advance
 A: Following @Qaiochu, if you know  (or better can show)  that $A_5$ is perfect,  then it's also a piece of cake.
That's $A_5'=A_5$.
Then if $H$ is any subgroup such that $A_5/H$ is abelian,  then $H\supset A_5'=A_5$.  So $H=A_5$.
A: As you have the simplicity of $A_5$ at hand, you can prove even more, namely that $A_5$ hasn't got subgroups of order $15$, $20$ and $30$.
In general, let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$: $$X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}.$$ If $G$ is simple, then:
$$[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$
In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.
So, $A_5$ can't have subgroups of order $15$, $20$ and $30$, because $(60/k)!<60$ for $k=15, 20, 30$.

$^\dagger$For $H\in X_G$, the kernel $K:=\bigcap_{g\in G}gHg^{-1}$ is a proper normal subgroup of $G$, and hence $K=\{1\}$ for the simplicity of $G$.
