simplicity of the alternating groups

Theorem 1.

If n5, then the alternating groupMathworldPlanetmath on n symbols, An, is simple.

Throughout the proof we extensively employ the cycle notation, with compositionMathworldPlanetmathPlanetmath on the left, as is usual. The symmetric groupMathworldPlanetmathPlanetmath on n symbols is denoted by Sn.

The following observation will be useful. Let π be a permutationMathworldPlanetmath written as disjoint cycles


It is easy to check that for every permutation σSn we have


As a consequence, two permutations of Sn are conjugatePlanetmathPlanetmath exactly when they have the same cycle type.

Two preliminary results will also be necessary.

Lemma 2.

The set of cycles of length 3 generates An.


A productPlanetmathPlanetmath of 3-cycles is an even permutationMathworldPlanetmath, so the subgroupMathworldPlanetmathPlanetmath generated by all 3-cycles is therefore contained in An. For the reverse inclusion, by definition every even permutation is the product of even numberMathworldPlanetmath of transpositionsMathworldPlanetmath. Thus, it suffices to show that the product of two transpositions can be written as a product of 3-cycles. There are two possibilities. Either the two transpositions move an element in common, say (a,b) and (a,c), or the two transpositions are disjoint, say (a,b) and (c,d). In the former case,


and in the latter,


This establishes the first lemma. ∎

Lemma 3.

If a normal subgroupMathworldPlanetmath NAn contains a 3-cycle, then N=An.


We will show that if (a,b,c)N, then the assumptionPlanetmathPlanetmath of normality implies that any other (a,b,c)N. This is easy to show, because there is some permutation in σSn that under conjugationMathworldPlanetmath takes (a,b,c) to (a,b,c), that is


In case σ is odd, then (because n5) we can choose some transposition (d,e)An disjoint from (a,b,c) so that


that is,


where σ is even. This means that N contains all 3-cycles, as NAn. Hence, by previous lemma N=An as required. ∎

Proof of theorem.

Let NAn be a non-trivial normal subgroup. We will show that N=An. The proof now proceeds by cases. In each case, the normality of N will allow us to reduce the proof to Lemma 2 or to one of the previous cases.

Case 1.

Suppose that there exists a πN that, when written as disjoint cycles, has a cycle of length at least 4, say


Upon conjugation by (a1,a2,a3)An, we obtain


Hence, πN, and hence ππ-1=(a1,a2,a4)N also. Notice that the rest of the cycles cancel. By Lemma 3, N=An.

Case 2.

Suppose that there exists a πN whose disjoint cycle decomposition has at least two cycles of length 3, say


Conjugation by (c,d,e)An implies that N also contains


Hence, N also contains ππ=(a,d,c,b,f). This reduces the proof to Case 1.

Case 3.

Suppose that there exists a πN whose disjoint cycle decomposition consists of exactly one 3-cycle and an even (possibly zero) number of transpositions. Hence, ππ is a 3-cycle. Lemma 3 can then be applied to completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Case 4.

Suppose there exists a πN of the form π=(a,b)(c,d). Conjugating by (a,e,b) with e distinct from a,b,c,d (at least one such e exists, as n5) yields


Hence ππ=(a,b,e)N. Again, Lemma 3 applies.

Case 5.

Suppose that N contains a permutation of the form


This time we conjugate by (a2,b1)(a3,b2).


Observe that


which reduces the proof to Case 2.

Since there exists at least one non-identity πN, and since this π is covered by one of the above cases, we conclude that N=An, as was to be shown.

Title simplicity of the alternating groups
Canonical name SimplicityOfTheAlternatingGroups
Date of creation 2013-03-22 13:07:57
Last modified on 2013-03-22 13:07:57
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 16
Author rmilson (146)
Entry type Result
Classification msc 20D06
Classification msc 20E32
Related topic ExamplesOfFiniteSimpleGroups