proof of simplicity of Mathieu groups
We give a uniform proof of the simplicity of the Mathieu groups , , and , and the alternating groups (for ), assuming the simplicity of and . (Essentially, we are assuming that the simplicity of the projective special linear groups is known.)
Lemma 1.
Let act transitively on a set . If is a normal subgroup of , then the transitivity classes of the action, restricted to , form a set of blocks for the action of .
Proof.
If , are any transitivity classes for the restricted action, let , , and such that . Then is a bijective map from onto (here we use normality). Hence any element of maps transitivity classes to transitivity classes. ∎
Hence it follows:
Corollary 2.
Let act primitively (http://planetmath.org/PrimativeTransitivePermutationGroupOnASet) on a set . If is a normal subgroup of , then either acts transitively on , or lies in the kernel of the action. If the action is faithful, then either or is transitive.
Theorem 3.
Let be a group acting primitively and faithfully on a set . Let be the stabilizer of some point , and assume that is simple. Then if is a nontrivial proper normal subgroup of , then is isomorphic to the semidirect product of by . can be identified with in such a way that is identified with , the action of becomes left multiplication, and the action of becomes conjugation.
Proof.
Since is a normal subgroup of , it is either or .
If , then , and since is maximal and is proper, we have . Since is normal and stabilizes , then stabilizes every point (since the action is transitive). Since the action is faithful, , a contradiction. (This contradiction can also be reached by applying the corollary.)
Therefore, . So no element of , other than 1, fixes . Thus acts freely and transitively on . For any , if and , then , hence is in . Thus is generated by and . Since is normal and , is the (internal) semidirect product of by . ∎
Now we come to the main theorem from which we will deduce the simplicity results.
Theorem 4.
Let be a group acting faithfully on a set . Let and let be the stabilizer of . Assume is simple.
1. Assume the action of is doubly transitive, and let be a nontrivial proper normal subgroup of . Then is an elementary abelian -group for some prime . Furthermore, is isomorphic to a subgroup of , and is isomorphic to a subgroup of , the group of affine transformations of .
2. If the action of is triply transitive and , then any nontrivial proper normal subgroup of is an elementary abelian 2-group.
3. If the action of is quadruply transitive and , then is simple.
Proof.
For part 1, use the identification of with given by the previous theorem. Since the action is doubly transitive, the action by conjugation of is transitive on . Therefore, all non-identity elements of have the same order, which must therefore be some prime . Hence is a -group. The center is nontrivial, and is preserved by all automorphisms. By double transitivity again, there is an automorphism taking any nontrivial element to any other; hence is abelian. Therefore is an elementary abelian -group.
For part 2, we know from part 1 that is isomorphic to an elementary abelian -group and acts as linear transformations of . Since the action of is triply transitive, the action of on the nonzero elements is doubly transitive. However, if , then the linearity of the action disallows double transitivity (if , then so we do not have complete freedom for any two elements since some element besides 0, and .)
(We note that when , we have the example , , .)
Here is an example illustrating part 2. The group acts triply transitively on , and the stabilizer of a point is , which is simple if . contains the normal subgroup of translations, an elementary abelian 2-group.
For part 3, note that the action of on , , is not triply transitive on nonzero elements, so the only conclusion left is that is simple. ∎
Corollary 5.
The Mathieu groups , , , and are simple.
Proof.
We take it as known that is simple. Since has as point stabilizer, and has a triply transitive action on a set of elements, we may work our way inductively up to , using the previous theorem. The index (http://planetmath.org/Coset) of in is , which is not a power of 2. Hence in all cases, is simple. ∎
Corollary 6.
The alternating groups are simple for .
Proof.
Since the natural action of on letters is quadruply transitive for , and the point stabilizer of is , we may apply the theorem to deduce the simplicity of the alternating groups , , from the simplicity of . ∎
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 |