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Math Help - Transitive action, blocks, primitive action, maximal subgroups.

  1. #1
    Senior Member abhishekkgp's Avatar
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    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.
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    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.
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    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.
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  4. #4
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    Re: Transitive action, blocks, primitive action, maximal subgroups.

    thank you! that was brilliant.
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