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A common proof of the simplicity of $A_5$ proceeds as follows.

First, note that a normal subgroup is always a union of conjugacy classes. Also, a subgroup has order dividing the size of the group by Lagrange's theorem. Any normal subgroup will contain the conjugacy class with the identity. Now if we look at unions of conjugacy classes of $A_5$ where we include $\{(1)\}$ in the union, there only two cases where the size of the union divides the order $A_5$. These are the cases where the union is all of $A_5$ or just $\{(1)\}$. Hence $A_5$ must be simple.

This same proof also works for $A_6$. But in general we cannot deduce the simplicity of $A_n$ for $n \geq 5$ with the same proof. According to this article it turns out that for example, $1 + \frac{n(n-1)(n-2)}{3}$ divides $n!/2$ when $n = 68$. Note that $\frac{n(n-1)(n-2)}{3}$ is the size of the conjugacy class of $3$-cycles in $A_n$ when $n \geq 5$.

Question: Is $n = 68$ the smallest positive integer where the proof fails? In the article they say that this "seems to be the smallest counterexample". However, counting all the possible sums of conjugacy classes of $A_n$ becomes computationally difficult very quickly (as is noted in the article). Hence actually demonstrating this might be hard.

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An optimization to the proof is probably well-known: If $N \unlhd A_n$, then $N \cap A_{n-1} \unlhd A_{n-1}$, so either $A_{n-1} \leq N$ (for all conjugates of $A_{n-1}$) or $A_{n-1} \cap N = 1$, so either $N=A_n$ or $N$ is quasi-regular and $|N| \leq n$. So one only needs to check very tiny conjugacy classes (and there are not any, since $A_n$ does not embed in $S_{n-1}$). –  Jack Schmidt Aug 7 '13 at 20:26
See also math.stackexchange.com/questions/402862/… where it is shown there are lots of solutions to the similar problem in Sn, and math.stackexchange.com/questions/41773/example-of-a-group/… where the proof fails in a smaller simple group. –  Jack Schmidt Aug 8 '13 at 14:13

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