# A group acts transitively on a finite set

Question: Assume the group G acts transitively on the finite set A and let $H\unlhd G$. If $\{\Theta_1 , ... , \Theta_n\}$ are the distinct orbits of H on A, prove that G acts transitively on these orbits.

Proposed Solution: Because H is normal, every $g\in G$ is contained in some coset of the quotient group, i.e. there exists $g',h'$ such that $g=g'h'$. Therefore, $g\cdot\Theta_i=\{g\cdot a_1,...,g\cdot a_n\}=\{g'h'\cdot a_1,...,g'h'\cdot a_n\}=\{g'\cdot a'_1,...,g'\cdot a'_n\}=g'\cdot\Theta_j$ and so $({g'}^{-1}g)\cdot\Theta_i=g''\cdot\Theta_i=\Theta_j$ which (I think) shows that all g send orbits to orbits. As for the transitivity, given $g\in G$, choosing 2 separate representatives of $gH$, i.e. $g_1h_1$ or $g_2h_2$ where $g_1h_1=g_2h_2=g$, and reproducing the above argument will be such that $({g_1}^{-1}g)\cdot\Theta_i\neq ({g_2}^{-1}g)\cdot\Theta_i$ and seeing as there are only finitely many orbits, I believe we are done.

Can somebody ensure my proof is correct?

• Comment on the first sentence of the proposed solution: The left $H$-cosets cover the group regardless of whether or not $H$ is normal. For example your decomposition could be written as $g=ge$ where $e$ is the group identity element. General comment: You introduce notation like $g',h',a_1,\ldots,a_n,\ldots$ without any hypotheses. You should explain each bit of notation you introduce when you introduce it. – ebrahim Nov 5 '15 at 21:11

As you understand, you first need to show that $G$ permutes the $H$-orbits, but I don't understand your first string of equalities.
Look, assume $a,b\in\Theta_i$ and $g.a\in\Theta_j$. You need to show $g.b\in\Theta_j$ as well. To do this, note that $b=h.a$ for some $h\in H$. Then, $$g.b=g.(h.a)=(gh).a=(ghg^{-1}).(g.a).$$ How do we conclude from this that $g.b\in \Theta_j$? (You know that $g.a\in\Theta_j$. What hypotheses on $H$ haven't been used yet?)
Next, you want the action of $G$ on the $H$-orbits to be transitive. Well, suppose $\Theta_i$ and $\Theta_j$ are two orbits. To show there is a $g\in G$ such that $g.\Theta_i=\Theta_j$, I can do the following:
Let $a\in\Theta_i$ and $b\in \Theta_j$. Then, there exists $g\in G$ such that $g.a=b$. (Why?) It follows that $g.\Theta_i=\Theta_j$. (Why?)