# Thread: Transitive action, blocks, primitive action, maximal subgroups.

1. ## Transitive action, blocks, primitive action, maximal subgroups.

Let $\displaystyle G$ act transitively on a finite set $\displaystyle A$. A 'block' in $\displaystyle A$ is a non-empty subset $\displaystyle B$ of $\displaystyle A$ such that for all $\displaystyle \sigma \in G$ either $\displaystyle \sigma(B)=B$ or $\displaystyle \sigma(B) \cap B= \phi$ (where $\displaystyle \sigma(B)=\{ \sigma(b)|b \in B \}$).
This action is called 'primitive' if the only blocks in $\displaystyle A$ are trivial ones: the sets of size $\displaystyle 1$ and $\displaystyle A$ itself.

Prove that:
The action(transitive) of $\displaystyle G$ on $\displaystyle A$ is primitive if and only if for each $\displaystyle a \in A,$ $\displaystyle G_a$ is a maximal subgroup of $\displaystyle G$. ($\displaystyle G_a=\{g \in G| g \cdot a=a \}=$ stabilizer of $\displaystyle a$ in $\displaystyle G$)

Here is what i have(with a little help from my friend):
Define $\displaystyle G_B= \{ \sigma \in G| \sigma(B)=B \}$. Its easy to see that $\displaystyle G_B \leq G$. Moreover if $\displaystyle a \in B$ then $\displaystyle G_a \leq G_B$.
I came to know that there exists a bijection between the blocks containing $\displaystyle a$ and the subgroups of $\displaystyle G$ containing $\displaystyle G_a$.I couldn't prove this. Help needed.

2. ## Re: Transitive action, blocks, primitive action, maximal subgroups.

First we show that if for each $\displaystyle a\in A$, $\displaystyle G_a$ is a maximal subgroup, then $\displaystyle A$ is primitive. In particular, we prove the contrapositive that if $\displaystyle G_a$ is not a maximal subgroup for some $\displaystyle a\in A$ then $\displaystyle A$ is not primitive. Let $\displaystyle G_a\leq H\leq G$ be a chain of proper subgroups. We claim that $\displaystyle H\cdot a=\{h\cdot a:h\in H\}$ is a block. It's easy to see that if $\displaystyle h\in H$ then $\displaystyle h\cdot(H\cdot a)=hH\cdot a=H\cdot a$. On the other hand, let $\displaystyle g\in G\setminus H$, and suppose towards a contradiction that $\displaystyle (g\cdot(H\cdot a))\cap (H\cdot a)\neq\emptyset$. Then there are $\displaystyle h_1,h_2\in H$ such that $\displaystyle gh_1\cdot a=h_2\cdot a$ and hence $\displaystyle (h_2)^{-1}gh_1\cdot a=a$. But then $\displaystyle (h_2)^{-1}gh_1\in G_a\leq H$, giving us $\displaystyle g\in H$, a contradiction. So $\displaystyle H\cdot a$ is indeed a block. Furthermore, $\displaystyle H\cdot a$ is nontrivial: for if $\displaystyle h\in H\setminus G_a$ then $\displaystyle h\cdot a$ and $\displaystyle a$ are distinct elements of $\displaystyle H\cdot a$. But $\displaystyle g\cdot a\notin H\cdot a$, because if it were then we would have $\displaystyle g\cdot a=h\cdot a$ for some $\displaystyle h\in H$ and hence $\displaystyle h^{-1}g\in G_a\leq H$, which would contradict the fact that $\displaystyle g\notin H$. Thus the first half of the proof is complete.

For the converse, let $\displaystyle B\subset A$ be a nontrivial block, and let $\displaystyle b,c\in B$ be distinct. Then $\displaystyle G_b\leq G_B\leq G$. We claim that both subgroup relations are proper: For let $\displaystyle a\in A\setminus B$. Then there is $\displaystyle g\in G$ with $\displaystyle a=g\cdot b\in g\cdot B$ by the transitivity of $\displaystyle A$, which means $\displaystyle g\notin G_B$. So $\displaystyle G_B$ is a proper subgroup of $\displaystyle G$. Also by the transitivity of $\displaystyle A$ there is $\displaystyle h\in G$ with $\displaystyle h\cdot b=c$, which means $\displaystyle h\notin G_b$. Since $\displaystyle B$ is a block with $\displaystyle (h\cdot B)\cap B\neq\emptyset$ then it must be that $\displaystyle h\cdot B=B$, and hence $\displaystyle h\in G_B$. So $\displaystyle G_b$ is a proper subgroup of $\displaystyle G_B$. We conclude that $\displaystyle G_b$ is not maximal, and this completes the second half of the proof.

3. ## Re: Transitive action, blocks, primitive action, maximal subgroups.

I was just thinking about how to show the stronger result that there is a bijection between blocks containing $\displaystyle a$ and the subgroups of $\displaystyle G$ containing $\displaystyle G_a$...

Let $\displaystyle \mathcal{B}$ as the set of blocks in $\displaystyle A$ containing $\displaystyle a$, and let $\displaystyle \mathcal{S}$ be the set of subgroups of $\displaystyle G$ containing $\displaystyle G_a$. Define $\displaystyle \varphi:\mathcal{B}\to\mathcal{S}$ by $\displaystyle \varphi(B)=G_B$, for each $\displaystyle B\in\mathcal{B}$. We claim that $\displaystyle \varphi$ is injective: let $\displaystyle B_1,B_2$ be distinct blocks containing $\displaystyle a$. Then (without loss of generality) there is $\displaystyle b_1\in B_1\setminus B_2$. By the transitivity of $\displaystyle \cdot$ there is $\displaystyle g\in G$ with $\displaystyle b_1=g\cdot a\in g\cdot B_2$. So $\displaystyle g\notin G_{B_2}$. However $\displaystyle a=g^{-1}\cdot b_1\in g^{-1}\cdot B_1$; since $\displaystyle B_1$ is a block this means $\displaystyle g^{-1}\cdot B_1=B_1$ and hence $\displaystyle g\in G_{B_1}$. Thus $\displaystyle G_{B_1}\neq G_{B_2}$, and it follows that $\displaystyle \varphi$ is injective.

Now we show that $\displaystyle \varphi$ is surjective: Let $\displaystyle G_a\leq H\leq G$ be a chain of subgroups of $\displaystyle G$. We showed in the previous post that $\displaystyle H\cdot a$ is a block, and obviously it contains $\displaystyle a$. We claim that $\displaystyle \varphi(H\cdot a)=H$. Clearly for each $\displaystyle h\in H$ we have $\displaystyle h\cdot(H\cdot a)=H\cdot a$; so $\displaystyle H\leq\varphi(H\cdot a)$. Now let $\displaystyle k\in\varphi(H\cdot a)$. Then $\displaystyle a\in H\cdot a=k\cdot(H\cdot a)$, which means there is $\displaystyle h\in H$ with $\displaystyle kh\cdot a=a$. So $\displaystyle kh\in G_a\leq H$, and thus $\displaystyle k\in H$. We conclude $\displaystyle H=\varphi(H\cdot a)$, and it follows that $\displaystyle \varphi$ is surjective, indeed, bijective.

4. ## Re: Transitive action, blocks, primitive action, maximal subgroups.

thank you! that was brilliant.