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Prove that $A_5$ is simple.

The solution presented is as follows:

Let $N$ be a non trivial normal subgroup of $A_5.$ If $(12)(34)\in N,$ then $$(25)(34)(12)(34)(34)(25) = (15)(34)$$ and $$(12)(34)(15)(34)=(152).$$ Again, if $(12345)\in N,$ then $$(23)(45)(12345)(45)(23) = (13254)$$ and $(12345)(13254) = (142).$ Now we know that, if a normal subgroup of $A_n$ contains even a single $3-$cycle it must be all of $A_n$, it follows that $N=A_5.$

In this solution I don't get, why only they consider these two cases $(12)(34)$ and $(12345),$ are present in $N.$ There are so many more cases possible for example, say if $(13)(24)$ may belong to $N$, then what? Obviously, we can verify, that we again find a three cycle in $A_5$, but I want to know how does the author make it so obvious that, there will be a three- cycle in $A_5$ for every other possible cases by just considering these two examples. As this seems not at all obvious to me. I dont get this at all...

I know there are tons of questions on this site concerning the same topic. But I want an explanation for this particular proof.

Nicky Hekster
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Arthur
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    It works the same for $(13)(24)\in N$. You just have to adapt the calculation. I am sure you can do it. Perhaps you are also interested in alternative proofs for the simplicity of $A_5$. – Dietrich Burde Feb 12 '23 at 13:21
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    The only other nontrivial cycle types besides $3$-cycles in $A_5$ are double transpositions or $5$-cycles. In both cases, we just check directly (for all of them, if you want, see the calculations), that then also a $3$-cycle is in $N$, so that $N=A_5$. So we are done. – Dietrich Burde Feb 12 '23 at 13:31
  • @DietrichBurde Ok. No problem. Thank you! I remain grateful – Arthur Feb 12 '23 at 13:43
  • You can conjugate $A_5$ by an appropriate $\sigma\in S_n$ that transforms $(13)(24)$ into $(12)(34)$; if $N$ was normal in $A_5$, so is the image under this conjugation. So if the image of $N$ equals all of $A_5$, then so does $N$ itself. In other words, you can just rename the elements being acted on and assume that you have $(12)(34)\in N$ if any product of two transpositions is in $N$. Similarly for a $5$-cycle. – Arturo Magidin Feb 13 '23 at 01:57

1 Answers1

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Any two permutations with the same cycle structure are conjugate in $S_n$.

So suppose your normal subgroup $N$ contains some permutation that is a product of two disjoint transpositions: $(ab)(cd)$. Then there is a permutation $\sigma\in S_5$ such that $\sigma(ab)(cd)\sigma^{-1}=(12)(34)$. Now consider the subgroup $\sigma N\sigma^{-1}\triangleleft \sigma A_5\sigma^{-1}=A_5$. If $\sigma N\sigma^{-1}=A_5$, then $N=A_5$. So considering the case of $(12)(34)$ covers all cases in which $N$ contains a permutation that is a product of two disjoint transpositions.

Likewise, if $N$ contains a $5$-cycle, then conjugating by a suitable $\sigma\in S_5$ you may assume that $N$ contains $(12345)$. Thus, it suffices to consder that case.

Arturo Magidin
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  • "If $ \sigma N\sigma^{-1}=A_5$ then $N=A_5$"- This is the part I don't understand . I mean did you try to imply " if $ \sigma N\sigma^{-1}=A_5$ then $N=A_5$" or $ \sigma N\sigma^{-1}=A_5$ is always true . Now, I know that if $N$ is a normal subgroup of $G$ then $gNg^{-1}$ is also a normal subgroup of $G$, $\forall g\in G$ and infact $gNg^{-1}=N$, but I couldn't understand it clrarly. Will you mind if I ask you for a bit more elucidation upon the part? – Arthur Feb 14 '23 at 11:44
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    @Franklin I meant exactly what I said. I said "if", I meant "if". If $H$ is a normal subgroup of $A$, and $A$ is a subgroup of $G$, then for every $g\in G$, $gHg^{-1}$ is a normal subgroup of $gAg^{-1}$. Do work that out for yourself in detail. Since $N$ is assumed to be normal in $A_5$, then $\sigma N\sigma^{-1}$ is normal in $\sigma A_5\sigma^{-1}$. But the latter is just $A_5$. If $\sigma N\sigma^{-1}=A_5$, then $N$ has the same finite number of elements as $A_5$, contained in $A_5$, hence equal to $A_5$. – Arturo Magidin Feb 14 '23 at 13:17
  • Thanks a ton, for responding, clarifying and enlightening me. Indeed the key to this approach was that :" If $H$ is a normal subgroup of $A$ and $A$ is a subgroup of $G$, then for every $g\in G,$ $gHg^{-1}$ is a normal subgroup of $gAg^{-1}$". Maybe, the author expects to "know this (most probably) a very popular (and obvious) result " to the readers. I get it now. Just another heuristical approach in solving these sort of problems, I stumbled upon! – Arthur Feb 14 '23 at 13:49