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I am new here and don't know much about Latex so, I attach my question from Permutation Groups by J. Dixon. I hope to get a help for it:

1.6.5 Let $G$ be a transitive subgroup of $\mathrm{Sym}(\Omega)$ and let $\alpha \in \Omega$. Show that $\mathrm{fix}(G_\alpha)$ is a block for $G$. In particular, if $G$ is primitive, then either $\mathrm{fix}(G_\alpha)=\{\alpha\}$ or else $G_\alpha=1$ and $G$ has prime degree.

Thank you.

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    A block "B" is a subset of "Omega" which for any "g in G", "B^g=B" or "B^g has no intersection with B but empty".2012-06-10

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A transitive permutation action comes from $G$ acting via right multiplication on $H=G_\alpha$. The fixed points $B=fix(G_\alpha)$ of $G_\alpha$ correspond to cosets of $H$ in $N_G(H)$. Saying that $g\in G$ is such that $gB\cap B\neq\emptyset$ is essentially the same as saying $(\alpha)g\in B$. This means $Hg\subset N_G(H)$, so that $g\in N_G(H)$; $g$ thus permutes the cosets of $H$ in $N_G(H)$, that is, fixes $B$ setwise.

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    Thanks so much for neat answer.2012-06-14