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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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Alec Teal
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  • $A_4$ is a subgroup but it's not a normal subgroup. $S_4$ is also not a normal subgroup of $S_5$. To play around with this, start with a nontrivial element of $S_5$, conjugate it a bunch of times, multiply everything you get in various ways, conjugate those things a bunch of times, and see what kind of elements you can get this way. – Qiaochu Yuan Jun 05 '14 at 19:15
  • @QiaochuYuan I realised that when reading again, thanks for pointing it out. – Alec Teal Jun 05 '14 at 19:17
  • @QiaochuYuan what is a good group theory text for problem solving oriented minds? – Asinomás Jun 05 '14 at 19:18

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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.

DonAntonio
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  • can u explain the fact "all transpositions are in the same conjugacy class" please ; i am not getting it – Learnmore Mar 13 '15 at 03:02
  • $S_5=HA_5$ does not necessarily lead to $|H|=2$. I think you need the 2nd isomorphism theorem here. – Michael Oct 26 '22 at 10:12
  • @Michael --- since these are finite groups, it suffices to count elements. Since $H\cap A_5 = {1}$, we know the order of $S_5 = HA_5$ is the product of the orders of $H$ and $A_5$. – Ben Mar 29 '23 at 14:30
  • Although we don't need finite groups. Let $G$ be a group with normal subgroup $H$ and $[G:H] = n$. Suppose $K\subset G$ is a subgroup with $K\cap H = {1}$ and $KH= G$. Then $K$ has order $n$. Proof: if $k_1,k_2\in K$ are such that $(k_2)^{-1}k_1\in H$ then $(k_2)^{-1} k_1 = 1$, so $k_1 = k_2$. That is, in $G/H$ (a group of order $n$), we have $k_1H = k_2H$ only if $k_1 = k_2$, showing the order of $K$ is at most $n$. On the other hand $KH = G$ implies $gH = khH = kH$ for some $k\in K, h\in H$, so every coset contains some representative $kH$, implying the order of $K$ is at most $n$. – Ben Mar 29 '23 at 14:52