Transitive action, blocks, primitive action, maximal subgroups.

**abhishekkgp** · August 6th 2011, 05:10 AM

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.

**hatsoff** · August 6th 2011, 12:33 PM

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.

**hatsoff** · August 6th 2011, 07:06 PM

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.

**abhishekkgp** · August 6th 2011, 07:39 PM

thank you! that was brilliant.

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Transitive action, blocks, primitive action, maximal subgroups.

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