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

• Aug 6th 2011, 05:10 AM
abhishekkgp
Transitive action, blocks, primitive action, maximal subgroups.
Let $G$ act transitively on a finite set $A$. A 'block' in $A$ is a non-empty subset $B$ of $A$ such that for all $\sigma \in G$ either $\sigma(B)=B$ or $\sigma(B) \cap B= \phi$ (where $\sigma(B)=\{ \sigma(b)|b \in B \}$).
This action is called 'primitive' if the only blocks in $A$ are trivial ones: the sets of size $1$ and $A$ itself.

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

Here is what i have(with a little help from my friend):
Define $G_B= \{ \sigma \in G| \sigma(B)=B \}$. Its easy to see that $G_B \leq G$. Moreover if $a \in B$ then $G_a \leq G_B$.
I came to know that there exists a bijection between the blocks containing $a$ and the subgroups of $G$ containing $G_a$.I couldn't prove this. Help needed.
• Aug 6th 2011, 12:33 PM
hatsoff
Re: Transitive action, blocks, primitive action, maximal subgroups.
First we show that if for each $a\in A$, $G_a$ is a maximal subgroup, then $A$ is primitive. In particular, we prove the contrapositive that if $G_a$ is not a maximal subgroup for some $a\in A$ then $A$ is not primitive. Let $G_a\leq H\leq G$ be a chain of proper subgroups. We claim that $H\cdot a=\{h\cdot a:h\in H\}$ is a block. It's easy to see that if $h\in H$ then $h\cdot(H\cdot a)=hH\cdot a=H\cdot a$. On the other hand, let $g\in G\setminus H$, and suppose towards a contradiction that $(g\cdot(H\cdot a))\cap (H\cdot a)\neq\emptyset$. Then there are $h_1,h_2\in H$ such that $gh_1\cdot a=h_2\cdot a$ and hence $(h_2)^{-1}gh_1\cdot a=a$. But then $(h_2)^{-1}gh_1\in G_a\leq H$, giving us $g\in H$, a contradiction. So $H\cdot a$ is indeed a block. Furthermore, $H\cdot a$ is nontrivial: for if $h\in H\setminus G_a$ then $h\cdot a$ and $a$ are distinct elements of $H\cdot a$. But $g\cdot a\notin H\cdot a$, because if it were then we would have $g\cdot a=h\cdot a$ for some $h\in H$ and hence $h^{-1}g\in G_a\leq H$, which would contradict the fact that $g\notin H$. Thus the first half of the proof is complete.

For the converse, let $B\subset A$ be a nontrivial block, and let $b,c\in B$ be distinct. Then $G_b\leq G_B\leq G$. We claim that both subgroup relations are proper: For let $a\in A\setminus B$. Then there is $g\in G$ with $a=g\cdot b\in g\cdot B$ by the transitivity of $A$, which means $g\notin G_B$. So $G_B$ is a proper subgroup of $G$. Also by the transitivity of $A$ there is $h\in G$ with $h\cdot b=c$, which means $h\notin G_b$. Since $B$ is a block with $(h\cdot B)\cap B\neq\emptyset$ then it must be that $h\cdot B=B$, and hence $h\in G_B$. So $G_b$ is a proper subgroup of $G_B$. We conclude that $G_b$ is not maximal, and this completes the second half of the proof.
• Aug 6th 2011, 07:06 PM
hatsoff
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 $a$ and the subgroups of $G$ containing $G_a$...

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

Now we show that $\varphi$ is surjective: Let $G_a\leq H\leq G$ be a chain of subgroups of $G$. We showed in the previous post that $H\cdot a$ is a block, and obviously it contains $a$. We claim that $\varphi(H\cdot a)=H$. Clearly for each $h\in H$ we have $h\cdot(H\cdot a)=H\cdot a$; so $H\leq\varphi(H\cdot a)$. Now let $k\in\varphi(H\cdot a)$. Then $a\in H\cdot a=k\cdot(H\cdot a)$, which means there is $h\in H$ with $kh\cdot a=a$. So $kh\in G_a\leq H$, and thus $k\in H$. We conclude $H=\varphi(H\cdot a)$, and it follows that $\varphi$ is surjective, indeed, bijective.
• Aug 6th 2011, 07:39 PM
abhishekkgp
Re: Transitive action, blocks, primitive action, maximal subgroups.
thank you! that was brilliant.