I can find proofs (but I admit I couldn't do them without referencing them) that $A_5$ is a normal subgroup of $S_5$, I also know (and the question says I may use - but without is more than welcome) that $A_5$ is simple, that is it's only normal subgroups are $\{1\}$ (identity) and $A_5$ itself.
I actually find that a bit hard to simply take, I shall look up a proof for this now. I think my difficulty stems from this.
Moving on, I cannot prove that if I have a normal subgroup of $S_5$ that is proper that it must be $A_5$
I have had some thoughts though. Consider $N$ and another normal subgroup $P$. If $P\ne N$ I'll want $|P\cap N|=1$ - I have no proof on what happens if this is not the case, I'm importing it based on $A_5$ being simple.
I've also tried to use: if I take an arbitrary odd permutation, $a$ that $aa$ is even, which makes that intersection greater than one. I now want to head torwards $P$ being the entire of $S_5$ but I'm not sure how.
I'd quite like to know how I should go about this. There are similar questions but right now Proving that $A_n$ is the only proper nontrivial normal subgroup of $S_n$, $n\geq 5$ is beyond me.
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$\begingroup$Some ideas:
$\;[S_5:A_5]=2\;$ so clearly $\;A_5\lhd S_5\;$ . About uniqueness:
$$H\lhd S_5\,\;\;H\neq 1,S_5\implies H\cap A_5\lhd A_5\implies H\cap A_5=1\;\;or\;\;H\cap A_5=A_5$$
because $\;A_5\;$ is simple, but:
$$\begin{align*}(1)&\;\;H\cap A_5=A_5\iff A_5\le H\implies |H|\ge 60\implies \begin{cases}H=S_5\\{}\\H=A_5\end{cases}\\{}\\(2)&\;\;H\cap A_5=1\implies |H|=2\;,\;\;\text{since}\;\;S_5=HA_5\end{align*}$$
But then $\;|H|=2\iff H=\langle \tau\rangle\;,\;\;\tau\;$ a single transposition (the only elements of order two that don't belong to $\;A_5\;$...), and since all the transpositions are in the same conjugacy class, all the transpositions are in $\;H\;$ , which is absurd.
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