Probably Random

A blog by Paul Siegel

The Alternating Group is Simple II

April 19, 2014

In my last post I described the alternating group and its place in the world of groups. I will now prove that A5 is simple, and in the third and final post of this series I will prove that An is simple for n5. The plan of attack is as follows: first I will carry out some preliminary analysis of conjugacy in Sn and An, and then by identifying all conjugacy classes in A5 I will prove that A5 is simple. I will then prove that An is simple for n5 by induction. I’m not sure who invented this argument; all I know is that I learned it in Dummit & Foote.

Conjugacy Classes in the Alternating Group

To understand the normal subgroups of a group it is very useful to first think carefully about its conjugacy classes; this is because a normal subgroup is by definition the union of conjugacy classes. Fortunately conjugation in the symmetric group is easy to understand using “cycle notation”. A k-cycle in Sn is a permutation which fixes all but k symbols a1,,ak which acts on these symbols as:

a1a2aka1

The notation for this cycle is (a1a2ak). It is not hard to show that every permutation decomposes as the product of disjoint cycles, and the decomposition is unique up to reordering the cycles. Indeed, cycle notation makes it particularly easy to understand conjugation.

Let σ=(a1a2ak) be a cycle and let τ be any permutation. Then τστ1=(τ(a1)τ(a2)τ(ak))

For i<k we have τστ1(τ(ai))=τσ(ai)=τ(ai+1) and similarly τστ1(τ(ak))=τ(a1).

The lemma extends easily to the case where σ is the product of cycles, so we see that conjugation by τ preserves the cycle structure of σ while relabelling the symbols in the cycle. In particular, two elements of Sn are conjugate if and only if the number and lengths of cycles are the same. For instance, (12)(345) is conjugate to (124)(35) in S5 but not to (12345).

Note that conjugacy in An is a little more subtle. A k-cycle is even if and only if k is odd, but not all k-cycles are conjugate in An. For instance the transposition τ=(45) conjugates (12345) to (12354) in S5, but there is no even permutation which conjugates (12345) to (12354) and hence they are not conjugate in A5.

To prove that A5 is simple, we will need to determine the sizes of all of its conjugacy classes. We will do this using the following tool:

Let g be an element of a group G, let ZG(g) be the centralizer of g (i.e. the set of all elements of G which commute with g) and let CG(g) denote the conjugacy class of g. Then

|ZG(g)||CG(g)|=|G|

Let G act on itself by conjugation. The orbit of g under this action is CG(g) and the stabilizer is ZG(g), so the result follows from the orbit-stabilizer theorem.

We will apply this lemma as follows. First we will use our understanding of conjugacy in Sn to identify the centralizer of a cycle. From that it is easy to identify the centralizer of a cycle in An, and that will allow us to count the conjugates of a cycle in An.

Let σSn be a k-cycle. Then: ZSn(σ)=σiτ::0i<k,τSnk

Proof: By Lemma 1, the conjugates of σ in Sn are precisely the k-cycles. To specify a k-cycle one must specify the symbols in the k-cycle and the order in which they appear; there are n!k!(nk)! ways to choose k symbols and k! different orders in which they can appear, though k of the orders define the same cyclic permutation. Thus there are n!k!(nk)!(k1)!=n!k(nk)! conjugates of σ; by Lemma 2, |ZSn(σ)|=k(nk)!.

The permutation σi clearly commutes with σ. Any permutation τ which fixes the k symbols that σ acts on also commutes with σ, and the subgroup of all such permutations is isomorphic to Snk. Thus the permutations σiτ, τSnk, all commute with σ; there are k(nk)! distinct permutations of this form, so they make up the entire centralizer of σ.

Simplicity of A5

We are now ready to prove the main result of this post:

A5 is simple.

The only possible cycle structures of non-identity elements in A5 are (123), (12345), and (12)(34). Recall that in S5 the cycle structure completely determines the conjugacy class; in A5 some of these conjugacy classes may split. Let us analyze each conjugacy class in turn using Proposition 1.

  • (123): The centralizer of (123) in S5 consists of the six permutations (123)iτ where i=0,1,2 and τ is either the identity or (45), so (123) has 120/6=20 conjugates in S5. If τ=(45) then (123)iτ is odd, so the centralizer in A5 has only three elements and hence the number of conjugates is still 60/3=20. Thus all 3-cycles are conjugate in A5.

  • (12345): The centralizer of (12345) in both S5 and A5 is just the cyclic subgroup (12345)i, so there are 120/5=24 conjugates in S5 and 60/5=12 conjugates in A5. The other 12 elements in the S5 conjugacy class are accounted for by the A5 conjugacy class of (12354) which is disjoint from that of (12345).

  • (12)(34): It is straightforward to check that (12)(34) commutes with the identity, itself, (13)(24) and (14)(23). If τ does not fix the symbol 5 then τ(12)(34)τ(12)(34) by Lemma 1, so (12)(34) does not commute with τ. A similar argument shows that (12)(34) does not commute with any 3-cycle, so the centralizer has exactly 4 elements and hence (12)(34) has 60/4=15 conjugates in A5.

Including the identity, we have accounted for the conjugacy classes of all 60 elements of A5: 60=1+20+12+12+15. So let H be a normal subgroup of A5. Since H is normal it is the union of conjugacy classes (including the identity), so |H| is the sum of 1 and some subset of 20,12,12,15. But |H| must also divide |A5|=60; checking cases the only possible choices for |H| are 1 and 60.