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I have a question on the argument in part (2) of the following theorem (it is a shortened version of this theorem from the book Permutation Groups by Dixon & Mortimer):

Theorem: Let $G$ be a group acting transitively on a set $\Omega$, and $H \unlhd G$. Then:

(1) the orbits of $H$ form a system of blocks for $G$;

(2) the group $H$ has at most $|G : H|$ orbits, and if the index $|G : H|$ is finite then the number of orbits of $H$ divides $|G : H|$;

(3) if $G$ acts primitively on $\Omega$ then either $H$ is transitive or $H$ lies in the kernel of the action.

The proof of (2) just notes:

This follows at once from (1) since all blocks in a system of blocks have the same size.

The proof (3) is a similar short note, but there it is totally clear to me, but for (2) why it could be readily seen by (1)? The arguments I found in the finite case are a little bit more involved: By orbit-stabilizer and as they all have the same size their common size is $|H : H_{\alpha}|$ for some $\alpha \in \Omega$ and $|\Omega| = k \cdot |H : H_{\alpha}|$ where $k$ denotes the number of orbits. Further as $G$ acts transitively $|\Omega| = |G : G_{\alpha}|$, and also we have $|G : H_{\alpha}| = |G:H|\cdot |H:H_{\alpha}|$ and with $H_{\alpha} \le G_{\alpha}$ also $|G:H_{\alpha}| = |G:G_{\alpha}||G_{\alpha}:H_{\alpha}|$. Putting all this together we have $$ \frac{|G:H_{\alpha}|}{|G_{\alpha}:H_{\alpha}|} = k \cdot \frac{|G:H_{\alpha}|}{|G:H|} $$ hence $|G:H| = k \cdot |G_{\alpha} : H_{\alpha}|$.

But the arguments above are just valid if the involved indices are finite, and besides being not simply seen at once, what is the simple argument that (2) follows at once from (1), and how is this applied in the infinite case?

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The orbits $\Delta=\Delta_1,\Delta_2, \ldots$ of $H$ form a block system, and there is an induced transitive action of $G$ on the set $\mathcal{B} = \{ \Delta_1,\Delta_2,\ldots\}$ of blocks, with $H$ contained in the stabilizer $G_\Delta$ of $\Delta$.

The total number $|\mathcal{B}|$ of blocks is the size of of the orbit of $\Delta$ under $G$ which, by the Orbit-Stabilizer Theorem is $|G:G_\Delta|$. Since $H \le G_{\Delta}$, this is at most $|G:H|$ and it divides $|G:H|$ if $|G:H|$ is finite.

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