proof of simplicity of Mathieu groups
We give a uniform proof of the simplicity of the Mathieu groups M22, M23, and M24, and the alternating groups An (for n>5), assuming the simplicity of M21≅PSL(3,𝔽4) and A5≅PSL(2,𝔽4). (Essentially, we are assuming that the simplicity of the projective special linear groups is known.)
Lemma 1.
Let G act transitively on a set S. If H is a normal subgroup of G, then the transitivity classes of the action, restricted to H, form a set of blocks for the action of G.
Proof.
If T, U are any transitivity classes for the restricted action, let t∈T, u∈U, and g∈G such that gt=u. Then x↦gx is a bijective 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 (http://planetmath.org/PrimativeTransitivePermutationGroupOnASet) 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 faithful, then either H={1} or H is transitive.
Theorem 3.
Let G be a group acting primitively and faithfully on a set S. Let K be the stabilizer of some point s0∈S, and assume that K is simple. Then if H is a nontrivial proper normal subgroup of G, then G is isomorphic to the semidirect product of H by K. H can be identified with S in such a way that 1∈H is identified with s0, the action of H becomes left multiplication, and the action of K becomes conjugation.
Proof.
Since H∩K is a normal subgroup of K, it is either {1} or K.
If H∩K=K, then K⊂H, 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 contradiction. (This contradiction can also be reached by applying the corollary.)
Therefore, H∩K={1}. So no element of H, other than 1, fixes s0. Thus H acts freely and transitively on S. For any g∈G, 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 H∩K={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 s0∈S 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 subgroup of GL(n,Fp), and G is isomorphic to a subgroup of AGL(n,Fp), the group of affine transformations 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.
Proof.
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 automorphisms. By double transitivity again, there is an automorphism taking any nontrivial element to any other; hence H is abelian. 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 x↦y, then 2x↦2y so we do not have complete 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 𝔽n2, 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 𝔽n2, n>2, is not triply transitive on nonzero elements, so the only conclusion left is that G is simple. ∎
Corollary 5.
The Mathieu groups M21, M22, M23, and M24 are simple.
Proof.
We take it as known that M21≅PSL(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 (http://planetmath.org/Coset) 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 n≥5.
Proof.
Since the natural action of An on n letters is quadruply transitive for n≥6, and the point stabilizer of An is An-1, we may apply the theorem to deduce the simplicity of the alternating groups An, n≥5, from the simplicity of A5≅PSL(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 |