# transitive actions are primitive if and only if stabilizers are maximal subgroups

###### Theorem 1.

If $G$ is transitive^{} on the set $A$, then $G$ is primitive on $A$ if and only if for each $a\mathrm{\in}A$, ${G}_{a}$ is a maximal subgroup of $G$. Here ${G}_{a}\mathrm{=}{\mathrm{Stab}}_{G}\mathit{}\mathrm{(}a\mathrm{)}$ is the stabilizer^{} of $a\mathrm{\in}A$.

###### Proof.

First claim that if $G$ is transitive on $A$ and $B\subset A$ is a block (http://planetmath.org/BlockSystem) with $a\in B$, then ${G}_{B}=\{\sigma \in G\mid \sigma (B)=B\}$ is a subgroup^{} of $G$ containing ${G}_{a}$. It is obvious that ${G}_{B}$ is a subgroup, since

$$\sigma \in {G}_{B}\Rightarrow \sigma (B)=B\Rightarrow {\sigma}^{-1}(\sigma (B))={\sigma}^{-1}(B)\Rightarrow B={\sigma}^{-1}(B)\Rightarrow {\sigma}^{-1}\in {G}_{B}$$ | ||

$$\sigma ,\tau \in {G}_{B}\Rightarrow (\sigma \tau )(B)=\sigma (\tau (B))=\sigma (B)=B\Rightarrow \sigma \tau \in {G}_{B}$$ |

But also, if $\sigma \in {G}_{a}$ for $a\in B$, then $\sigma (a)=a$, so $\sigma (B)\cap B\ne \mathrm{\varnothing}$ and thus $\sigma (B)=B$ since $B$ is a block system and thus $\sigma \in {G}_{B}$. This proves the claim.

To prove the theorem, note that for each $a\in A$, there is by the claim a $1-1$ correspondence between containing $a$ and subgroups of $G$ containing ${G}_{a}$. Thus, $G$ is primitive on $A$ if and only if all blocks are either of size $1$ or equal to $A$, if and only if any group containing ${G}_{a}$ is either ${G}_{a}$ itself or $G$, if and only if for all $a\in A$, ${G}_{a}$ is maximal in $G$. ∎

Title | transitive actions are primitive if and only if stabilizers are maximal subgroups |
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Canonical name | TransitiveActionsArePrimitiveIfAndOnlyIfStabilizersAreMaximalSubgroups |

Date of creation | 2013-03-22 17:19:07 |

Last modified on | 2013-03-22 17:19:07 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 20B15 |