I am teaching an introductory group theory course, and it has come to the inevitable proof that $A_n$ is simple for $n\geq 5$. Now, there seem to be a number of proofs that I can find – one the "standard" one with $3$cycles, and the others using primitivity or conjugacy class size estimation. Does anyone have a list of proofs of simplicity of $A_n$ with comments? I am sure there are a number of different ideas which could be used... In particular, any method that generalizes to some of the other finite simple groups would be of particular interest...

1$\begingroup$ Using Iwasawa's Criterion would be a good thing to do if you're thinking of generalizing to other FSG's (it's good for all the classicals). $\endgroup$ – Nick Gill Mar 9 '16 at 14:07

$\begingroup$ compute the character table, and see that only the trivial character has a kernel. $\endgroup$ – Dima Pasechnik Mar 9 '16 at 14:23

1$\begingroup$ ... Iwasawa's Criterion will also do all the other alternating groups if you consider the action on 3subsets. (Actually, rather than say all the classicals, perhaps I should hedge my bets and exclude the orthogonals  I can't immediately remember the way the proof goes there.) I wrote a bunch of notes on this by the way that I am happy to send you (although it's all very classical). $\endgroup$ – Nick Gill Mar 9 '16 at 14:23

5$\begingroup$ This document has FIVE proofs of the simplicity of $A_n$, for $n\geq 5$: math.uconn.edu/~kconrad/blurbs/grouptheory/Ansimple.pdf $\endgroup$ – Nick Gill Mar 9 '16 at 14:54

1$\begingroup$ Incidentally for generals prep I'd wondered about this, but I don't think this is at all what you want: by MurnaghanNakayama, the only normal subgroups of S_n, n\geq 5, are the evident ones. Hence if H in A_n is normal it suff. to show that H\cap tHt^{1} is nontrivial (t a transp.). Thus it suff. to find an element of H in a conj. class that doesn't split  i.e. isn't a product of odd cycles of distinct lengths. Taking any such element, we're done (by taking powers) unless wlog it's (123...p), p\neq 3. Multiplying it with its conjugate under (12)(34) we get a (p2)cycle, QED by induction. $\endgroup$ – alpoge Mar 28 '16 at 12:43
Just turning comments into an answer.
Iwasawa's Criterion. This will do $A_5$ (using the natural action), $A_n$ with $n> 6$ (by considering the action on the set of all 3subsets of $\{1,\dots, n\}$) and a whole bunch of the classical groups (using transvections acting on the associated polar space). Some notes that might be useful are here.
A list of the type you asked for does exist thanks to Keith Conrad (it contains five proofs of the simplicity of $A_n$, for $n\geq 5$.)
See the comments for other suggested proofs.

1$\begingroup$ Isn't the action of $A_6$ on 3element subsets imprimitive because a 3set and its complement form a block? $\endgroup$ – Benjamin Steinberg Apr 26 at 20:23

$\begingroup$ @BenjaminSteinberg, hmmm, yes, $A_6$ does seem to need something else... BTW, I realised that the first link is broken, so I fixed it. $\endgroup$ – Nick Gill Apr 27 at 7:44
One inductive proof could go as follows: it is remarked in J. Gallian's "Introduction to Abstract Algebra" text (maybe as an exercise) that one way to prove that $A_{5}$ is simple is to note that every normal subgroup of a finite group $G$ is a union of conjugacy classes which must include $\{1_{G} \}$. Since the conjugacy class sizes for $A_{5}$ are $1,15,20,12$ and $12$ it is easy to check that $\{1_{G}\}$ and $G$ are the only normal subgroups.
Now assume that $n > 5$ and that $A_{k}$ is simple for $5 \leq k < n$. Let $M$ be a proper nonidentity normal subgroup of $G = A_{n}$. Let $H_{i}$ be the stabilizer of the point $i$. Then $H_{i} \cong A_{n1}$ is simple so that $M \cap H_{i} = \{ 1_{G} \}$. Now $G/N$ contains a subgroup isomorphic to $H_{i}$, so that $[G:M] \geq (n1)!$ and $M \leq n$. On the other hand, $G$ is doubly transitive, so that $H_{i}$ is a maximal subgroup of $G$ and thus $H_{i}M = G$. Hence $M = n$ and $M$ is a regular normal subgroup of $G$. Also, $M$ is a maximal normal subgroup, since $G/M$ is simple.
Now $M$ contains no $3$cycle, and $C_{G}(M) \lhd G$, so there is a $3$cycle $\sigma$ and an element $m \in M$ with $\sigma^{1}m^{1}\sigma m \neq 1$.
But $\sigma^{1}m^{1}\sigma m \in M$ and $\sigma^{1}m^{1}\sigma m $ is a product of two $3$cycles, so moves at most $6$ points. Hence $n \leq 6$, so in fact we must have $n=6$. However, since $H_{i}$ is a maximal subgroup of $G$, $M$ must be a minimal normal subgroup of $G$. But any group of order $6$ has a normal Sylow $3$subgroup by Sylow's Theorem. Thus the unique Sylow $3$subgroup of $M$ is normal in $G$, a contradiction.

$\begingroup$ I suppose it would be slicker to prove first that $A_{5}$ and $A_{6}$ are simple, and then prove that for $n \geq 7$, any proper nonidentity normal subgroup of $A_{n}$ contains a nonidentity element which fixes a point which then does it by induction. However the proof above illustrates more general grouptheoretic ideas . $\endgroup$ – Geoff Robinson Mar 11 '16 at 9:03
I prefer the proof going via 3cycles. It is probably the least elegant proof, but it is the one which you probably would have found when considering the problem without any prior knowledge. Also the general strategy generalizes to classical groups as well as Lie groups: Look for elements $g_1$, which cannot be contained in a normal subgroup, then look for elements $g_2$, such that the normal subgroup generated by $g_2$ contains $g_1$, and continue until you run out of elements to check. The disadvantage is that for each new class of groups you have to come up with the right elements, e.g. 3cycles for $A_n$, transvections for $Sl_n$, but if you choose the element to start with "close to the identity", you will succeed quite often.