proof of simplicity of Mathieu groups

We give a uniform proof of the simplicity of the Mathieu groupsMathworldPlanetmath M22, M23, and M24, and the alternating groupsMathworldPlanetmath An (for n>5), assuming the simplicity of M21PSL(3,𝔽4) and A5PSL(2,𝔽4). (Essentially, we are assuming that the simplicity of the projective special linear groupsMathworldPlanetmath is known.)

Lemma 1.

Let G act transitively on a set S. If H is a normal subgroupMathworldPlanetmath of G, then the transitivity classes of the action, restricted to H, form a set of blocks for the action of G.


If T, U are any transitivity classes for the restricted action, let tT, uU, and gG such that gt=u. Then xgx is a bijectiveMathworldPlanetmathPlanetmath map from T onto U (here we use normality). Hence any element of G maps transitivity classes to transitivity classes. ∎

Hence it follows:

Corollary 2.

Let G act primitively ( on a set S. If H is a normal subgroup of G, then either H acts transitively on S, or H lies in the kernel of the action. If the action is faithfulPlanetmathPlanetmath, then either H={1} or H is transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

Theorem 3.

Let G be a group acting primitively and faithfully on a set S. Let K be the stabilizerMathworldPlanetmath of some point s0S, and assume that K is simple. Then if H is a nontrivial proper normal subgroup of G, then G is isomorphicPlanetmathPlanetmathPlanetmathPlanetmath to the semidirect productMathworldPlanetmath of H by K. H can be identified with S in such a way that 1H is identified with s0, the action of H becomes left multiplication, and the action of K becomes conjugationMathworldPlanetmath.


Since HK is a normal subgroup of K, it is either {1} or K.

If HK=K, then KH, and since K is maximal and H is proper, we have K=H. Since H is normal and H stabilizes s0, then H stabilizes every point (since the action is transitive). Since the action is faithful, K=H={1}, a contradictionMathworldPlanetmathPlanetmath. (This contradiction can also be reached by applying the corollary.)

Therefore, HK={1}. So no element of H, other than 1, fixes s0. Thus H acts freely and transitively on S. For any gG, if gs0=s and hs=s0, then hgs0=s0, hence hg is in K. Thus G is generated by H and K. Since H is normal and HK={1}, G is the (internal) semidirect product of H by K. ∎

Now we come to the main theorem from which we will deduce the simplicity results.

Theorem 4.

Let G be a group acting faithfully on a set S. Let s0S and let K be the stabilizer of s0. Assume K is simple.

1. Assume the action of G is doubly transitive, and let H be a nontrivial proper normal subgroup of G. Then H is an elementary abelian p-group for some prime p. Furthermore, K is isomorphic to a subgroupMathworldPlanetmathPlanetmath of GL(n,Fp), and G is isomorphic to a subgroup of AGL(n,Fp), the group of affine transformationsMathworldPlanetmathPlanetmath of H.

2. If the action of G is triply transitive and |S|>3, then any nontrivial proper normal subgroup of G is an elementary abelian 2-group.

3. If the action of G is quadruply transitive and |S|>4, then G is simple.


For part 1, use the identification of H with S given by the previous theorem. Since the action is doubly transitive, the action by conjugation of K is transitive on H-{1}. Therefore, all non-identity elements of H have the same order, which must therefore be some prime p. Hence H is a p-group. The center Z(H) is nontrivial, and is preserved by all automorphismsMathworldPlanetmathPlanetmathPlanetmathPlanetmath. By double transitivity again, there is an automorphism taking any nontrivial element to any other; hence H is abelianMathworldPlanetmath. Therefore H is an elementary abelian p-group.

For part 2, we know from part 1 that H is isomorphic to an elementary abelian p-group and K acts as linear transformations of H. Since the action of G is triply transitive, the action of K on the nonzero elements is doubly transitive. However, if p>2, then the linearity of the action disallows double transitivity (if xy, then 2x2y so we do not have completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath freedom for any two elements since H some element besides 0, y and 2y.)

(We note that when |S|=3, we have the example G=S3, H=A3, K=S2.)

Here is an example illustrating part 2. The group AGL(n,𝔽2) acts triply transitively on 𝔽2n, and the stabilizer of a point is GL(n,𝔽2), which is simple if n>2. AGL(n,𝔽2) contains the normal subgroup of translations, an elementary abelian 2-group.

For part 3, note that the action of GL(n,𝔽2) on 𝔽2n, n>2, is not triply transitive on nonzero elements, so the only conclusionMathworldPlanetmath left is that G is simple. ∎

Corollary 5.

The Mathieu groups M21, M22, M23, and M24 are simple.


We take it as known that M21PSL(3,𝔽4) is simple. Since Mn has Mn-1 as point stabilizer, and has a triply transitive action on a set of n elements, we may work our way inductively up to M24, using the previous theorem. The index ( of Mn-1 in Mn is n, which is not a power of 2. Hence in all cases, Mn is simple. ∎

Corollary 6.

The alternating groups An are simple for n5.


Since the natural action of An on n letters is quadruply transitive for n6, and the point stabilizer of An is An-1, we may apply the theorem to deduce the simplicity of the alternating groups An, n5, from the simplicity of A5PSL(2,𝔽4)PSL(2,𝔽5). ∎

Title proof of simplicity of Mathieu groups
Canonical name ProofOfSimplicityOfMathieuGroups
Date of creation 2013-03-22 18:44:08
Last modified on 2013-03-22 18:44:08
Owner monster (22721)
Last modified by monster (22721)
Numerical id 8
Author monster (22721)
Entry type Proof
Classification msc 20D08
Classification msc 20B20