proof of simplicity of Mathieu groups

We give a uniform proof of the simplicity of the Mathieu groups $M_{22}$, $M_{23}$, and $M_{24}$, and the alternating groups $A_{n}$ (for $n>5$), assuming the simplicity of $M_{21}\cong PSL(3,\mathbb{F}_{4})$ and $A_{5}\cong PSL(2,\mathbb{F}_{4})$. (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 $t\in T$, $u\in U$, and $g\in G$ such that $gt=u$. Then $x\mapsto gx$ 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 $s_{0}\in S$, 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 $1\in H$ is identified with $s_{0}$, the action of $H$ becomes left multiplication, and the action of $K$ becomes conjugation.

Proof.

Since $H\cap K$ is a normal subgroup of $K$, it is either $\{1\}$ or $K$.

If $H\cap K=K$, then $K\subset H$, and since $K$ is maximal and $H$ is proper, we have $K=H$. Since $H$ is normal and $H$ stabilizes $s_{0}$, 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, $H\cap K=\{1\}$. So no element of $H$, other than 1, fixes $s_{0}$. Thus $H$ acts freely and transitively on $S$. For any $g\in G$, if $gs_{0}=s$ and $hs=s_{0}$, then $hgs_{0}=s_{0}$, hence $hg$ is in $K$. Thus $G$ is generated by $H$ and $K$. Since $H$ is normal and $H\cap K=\{1\}$, $G$ 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 $s_{0}\in S$ and let $K$ be the stabilizer of $s_{0}$. 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 $GL(n,\mathbb{F}_{p})$, and $G$ is isomorphic to a subgroup of $AGL(n,\mathbb{F}_{p})$, 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 $x\mapsto y$, then $2x\mapsto 2y$ 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 $G=S_{3}$, $H=A_{3}$, $K=S_{2}$.)

Here is an example illustrating part 2. The group $AGL(n,\mathbb{F}_{2})$ acts triply transitively on $\mathbb{F}_{2}^{n}$, and the stabilizer of a point is $GL(n,\mathbb{F}_{2})$, which is simple if $n>2$. $AGL(n,\mathbb{F}_{2})$ contains the normal subgroup of translations, an elementary abelian 2-group.

For part 3, note that the action of $GL(n,\mathbb{F}_{2})$ on $\mathbb{F}_{2}^{n}$, $n>2$, is not triply transitive on nonzero elements, so the only conclusion left is that $G$ is simple. ∎

Corollary 5.

The Mathieu groups $M_{21}$, $M_{22}$, $M_{23}$, and $M_{24}$ are simple.

Proof.

We take it as known that $M_{21}\cong PSL(3,\mathbb{F}_{4})$ is simple. Since $M_{n}$ has $M_{n-1}$ as point stabilizer, and has a triply transitive action on a set of $n$ elements, we may work our way inductively up to $M_{24}$, using the previous theorem. The index (http://planetmath.org/Coset) of $M_{n-1}$ in $M_{n}$ is $n$, which is not a power of 2. Hence in all cases, $M_{n}$ is simple. ∎

Corollary 6.

The alternating groups $A_{n}$ are simple for $n\geq 5$.

Proof.

Since the natural action of $A_{n}$ on $n$ letters is quadruply transitive for $n\geq 6$, and the point stabilizer of $A_{n}$ is $A_{n-1}$, we may apply the theorem to deduce the simplicity of the alternating groups $A_{n}$, $n\geq 5$, from the simplicity of $A_{5}\cong PSL(2,\mathbb{F}_{4})\cong PSL(2,\mathbb{F}_{5})$. ∎

Title proof of simplicity of Mathieu groups ProofOfSimplicityOfMathieuGroups 2013-03-22 18:44:08 2013-03-22 18:44:08 monster (22721) monster (22721) 8 monster (22721) Proof msc 20D08 msc 20B20