In my last post I described the alternating group and its place in the world of groups. I will now prove that A_5 is simple, and in the third and final post of this series I will prove that A_n is simple for n \geq 5. The plan of attack is as follows: first I will carry out some preliminary analysis of conjugacy in S_n and A_n, and then by identifying all conjugacy classes in A_5 I will prove that A_5 is simple. I will then prove that A_n is simple for n \geq 5 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 S_n is a permutation which fixes all but k symbols a_1, \ldots, a_k which acts on these symbols as:

a_1 \to a_2 \to \ldots \to a_k \to a_1

The notation for this cycle is (a_1 a_2 \ldots a_k). 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.

Lemma 1: Let \sigma = (a_1 a_2 \ldots a_k) be a cycle and let \tau be any permutation. Then \tau \sigma \tau^{-1} = (\tau(a_1) \tau(a_2) \ldots \tau(a_k))

Proof: For i < k we have \tau \sigma \tau^{-1}(\tau(a_i)) = \tau \sigma(a_i) = \tau(a_{i+1}) and similarly \tau \sigma \tau^{-1}(\tau(a_k)) = \tau(a_1).
QED

The lemma extends easily to the case where \sigma is the product of cycles, so we see that conjugation by \tau preserves the cycle structure of \sigma while relabelling the symbols in the cycle. In particular, two elements of S_n 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 S_5 but not to (12345).

Note that conjugacy in A_n 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 A_n. For instance the transposition \tau = (45) conjugates (12345) to (12354) in S_5, but there is no even permutation which conjugates (12345) to (12354) and hence they are not conjugate in A_5.

To prove that A_5 is simple, we will need to determine the sizes of all of its conjugacy classes. We will do this using the following tool:

Lemma 2: Let g be an element of a group G, let Z_G(g) be the centralizer of g (i.e. the set of all elements of G which commute with g) and let C_G(g) denote the conjugacy class of g. Then |Z_G(g)| \cdot |C_G(g)| = |G|

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

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

Proposition 3: Let \sigma \in S_n be a k-cycle. Then:
Z_{S_n}(\sigma) = \{\sigma^i \tau:\: 0 \leq i < k,\, \tau \in S_{n-k}\}

Proof: By Lemma 1, the conjugates of \sigma in S_n 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 \frac{n!}{k!(n-k)!} 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 \frac{n!}{k!(n-k)!} \cdot (k-1)! = \frac{n!}{k \cdot (n-k)!} conjugates of \sigma; by Lemma 2, |Z_{S_n}(\sigma)| = k \cdot (n-k)!.

The permutation \sigma^i clearly commutes with \sigma. Any permutation \tau which fixes the k symbols that \sigma acts on also commutes with \sigma, and the subgroup of all such permutations is isomorphic to S_{n-k}. Thus the permutations \sigma^i \tau, \tau \in S_{n-k}, all commute with \sigma; there are k \cdot (n-k)! distinct permutations of this form, so they make up the entire centralizer of \sigma.
QED

Simplicity of A_5

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

Theorem: A_5 is simple.

Proof: The only possible cycle structures of non-identity elements in A_5 are (123), (12345), and (12)(34). Recall that in S_5 the cycle structure completely determines the conjugacy class; in A_5 some of these conjugacy classes may split. Let us analyze each conjugacy class in turn using Proposition 3.

  • (123): The centralizer of (123) in S_5 consists of the six permutations (123)^i \tau where i = 0, 1, 2 and \tau is either the identity or (45), so (123) has 120/6 = 20 conjugates in S_5. If \tau = (45) then (123)^i \tau is odd, so the centralizer in A_5 has only three elements and hence the number of conjugates is still 60/3 = 20. Thus all 3-cycles are conjugate in A_5.
  • (12345): The centralizer of (12345) in both S_5 and A_5 is just the cyclic subgroup \{(12345)^i\}, so there are 120/5 = 24 conjugates in S_5 and 60/5 = 12 conjugates in A_5. The other 12 elements in the S_5 conjugacy class are accounted for by the A_5 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 \tau does not fix the symbol 5 then \tau (12)(34) \tau \neq (12)(34) by Lemma 1, so (12)(34) does not commute with \tau. 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 A_5.

Including the identity, we have accounted for the conjugacy classes of all 60 elements of A_5: 60 = 1 + 20 + 12 + 12 + 15. So let H be a normal subgroup of A_5. 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 |A_5| = 60; checking cases the only possible choices for |H| are 1 and 60.
QED

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