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I am trying to prove the simplicity of $A_5$ by showing that every non-trivial normal subgroup $H$ contains a 3-cycle, and therefore is all of $A_5$ since the 3-cycles all belong to one conjugacy class and all the 3-cycles generate $A_5$. Now if $H$ contains a 3-cycle already then the proof is done. Otherwise $H$ contains only products of two transpositions or 5-cycles. How can I cook up some general calculation that no matter what I'll always get a 3-cycle from these types of elements?

I've asked a related question here.

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If $H$ contains a 5-cycle then it contains all 5-cycles, since all subgroups of order 5 are conjugate. Show that the product of two well-chosen 5-cycles gives a 3-cycle.

If $H$ contains a permutation that is a product of two disjoint 2-cycles, then $H$ contains all such permutations (all are conjugate). Again, pick two and take their product.

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    $\begingroup$ This doesn't really affect the validity of the argument, but it is not true that all $5$-cycles are conjugate in $A_5$. But all subgroups of order $5$ are conjugate, so any normal subgroup containing a $5$-cycle contains all $5$-cycles. $\endgroup$ – Derek Holt May 7 '16 at 10:04
  • $\begingroup$ @DerekHolt Thank you for pointing this out, I posted this very quickly without thinking too deeply apparently. $\endgroup$ – Morgan Rodgers May 7 '16 at 10:48
  • $\begingroup$ When you say that a subgroup is conjugate you mean it is normal? $\endgroup$ – Jacopo Stifani May 8 '16 at 9:34
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    $\begingroup$ @JacopoStifani No, two subgroups $A$ and $B$ are conjugate if $gAg^{-1} = B$ for some $g \in G$. If a subgroup $A$ is normal then it is conjugate only to itself, so $gAg^{-1} = A$ for all $g \in G$. $\endgroup$ – Morgan Rodgers May 8 '16 at 9:37
  • $\begingroup$ And how do we know all subgroups of order 5 are conjugate? And why must a normal subgroup contain all 5-cycles? $\endgroup$ – Jacopo Stifani May 8 '16 at 16:30
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A very nice proof is the following, due to Galois. Let $N$ be a normal subgroup, and suppose that it is nontrivial. Pick an element $\sigma$ with the maximum number of fixed points that is not the identity. If it is not a three cycle, its cycle decomposition is of the form $\sigma =(123\cdots)\cdots$ or $\sigma =(12)(34)\cdots$. If you think about it for a while, you should be able to conjugate $\sigma$ appropriately so it fixes more elements, which contradicts our choice of $\sigma$.

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