# 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 $H\mathrm{=}\mathrm{\{}\mathrm{1}\mathrm{\}}$ 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 ${s}_{\mathrm{0}}\mathrm{\in}S$, 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 $\mathrm{1}\mathrm{\in}H$ is identified with ${s}_{\mathrm{0}}$, 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 $g{s}_{0}=s$ and $hs={s}_{0}$, then $hg{s}_{0}={s}_{0}$, 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 ${s}_{\mathrm{0}}\mathrm{\in}S$ and let $K$ be the stabilizer of ${s}_{\mathrm{0}}$. 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 $G\mathit{}L\mathit{}\mathrm{(}n\mathrm{,}{\mathrm{F}}_{p}\mathrm{)}$, and $G$ is isomorphic to a subgroup of $A\mathit{}G\mathit{}L\mathit{}\mathrm{(}n\mathrm{,}{\mathrm{F}}_{p}\mathrm{)}$, the group of affine transformations^{} of $H$.

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

3. If the action of $G$ is quadruply transitive and $\mathrm{|}S\mathrm{|}\mathrm{>}\mathrm{4}$, 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 ${M}_{\mathrm{21}}$, ${M}_{\mathrm{22}}$, ${M}_{\mathrm{23}}$, and ${M}_{\mathrm{24}}$ 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 $n\mathrm{\ge}\mathrm{5}$.

###### Proof.

Since the natural action of ${A}_{n}$ on $n$ letters is quadruply transitive for $n\ge 6$, and the point stabilizer of ${A}_{n}$ is ${A}_{n-1}$, we may apply the theorem to deduce the simplicity of the alternating groups ${A}_{n}$, $n\ge 5$, 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 |
---|---|

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 |