Simplicity of alternating group $A_n$

Question

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...

Answer 1

Just turning comments into an answer.

1. 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 3-subsets 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.

2. 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 $n\geq 5$.)

3. See the comments for other suggested proofs.

Answer 2

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 non-identity normal subgroup of $G = A_{n}$. Let $H_{i}$ be the stabilizer of the point $i$. Then $H_{i} \cong A_{n-1}$ is simple so that $M \cap H_{i} = \{ 1_{G} \}$. Now $G/N$ contains a subgroup isomorphic to $H_{i}$, so that $[G:M] \geq (n-1)!$ 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.

Answer 3

I prefer the proof going via 3-cycles. 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. 3-cycles 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.