1. Group Action

Let G be a finite group acting on the set S. Suppose $\displaystyle H\trianglelefteq G$ so that for any $\displaystyle s_1,s_2\in S$ there is a unique $\displaystyle h\in H$ so that $\displaystyle hs_1=s_2$. For each $\displaystyle s\in S$, let $\displaystyle G_s=\{g\in G: gs=s\}$.

Prove:

$\displaystyle G=G_sH \ \text{and} \ G_s\cap H=\{e\}$

I think about trying to use the order since if $\displaystyle |G|=|G_sH|=\frac{|G_s||H|}{|H\cap G_s|}$ but other than that I don't know what I need to do.

2. Re: Group Action

Originally Posted by dwsmith
Let G be a finite group acting on the set S. Suppose $\displaystyle H\trianglelefteq G$ so that for any $\displaystyle s_1,s_2\in S$ there is a unique $\displaystyle h\in H$ so that $\displaystyle hs_1=s_2$. For each $\displaystyle s\in S$, let $\displaystyle G_s=\{g\in G: gs=s\}$.

Prove:

$\displaystyle G=G_sH \ \text{and} \ G_s\cap H=\{e\}$

I think about trying to use the order since if $\displaystyle |G|=|G_sH|=\frac{|G_s||H|}{|H\cap G_s|}$ but other than that I don't know what I need to do.
You made the correct observation. If we can prove that $\displaystyle \text{stab}(s)\cap H=\{e\}$ then $\displaystyle |\text{stab}(s)H|=|G|$ and so $\displaystyle \text{stab}(s)H=G$. Now, suppose for a second that $\displaystyle h\in \text{stab}(s)\cap H$ then you know $\displaystyle hs=s$ but you know that there is a UNIQUE $\displaystyle h$ which does this. What could this $\displaystyle h$ be?

3. Re: Group Action

Originally Posted by Drexel28
You made the correct observation. If we can prove that $\displaystyle \text{stab}(s)\cap H=\{e\}$ then $\displaystyle |\text{stab}(s)H|=|G|$ and so $\displaystyle \text{stab}(s)H=G$. Now, suppose for a second that $\displaystyle h\in \text{stab}(s)\cap H$ then you know $\displaystyle hs=s$ but you know that there is a UNIQUE $\displaystyle h$ which does this. What could this $\displaystyle h$ be?
You have $\displaystyle hs=s$ which is true if $\displaystyle h\in G_s$ but the unique h is for $\displaystyle hs_1=s_2$ where $\displaystyle s_1 \ \text{and} \ s_2$ aren't necessarily the same.

4. Re: Group Action

Originally Posted by dwsmith
You have $\displaystyle hs=s$ which is true if $\displaystyle h\in G_s$ but the unique h is for $\displaystyle hs_1=s_2$ where $\displaystyle s_1 \ \text{and} \ s_2$ aren't necessarily the same.
You know that FOR ANY two elements of $\displaystyle S$ there exists a UNIQUE element of $\displaystyle H$ sending one to the other. What always fixes $\displaystyle s$, and how does that help us?

5. Re: Group Action

but IF $\displaystyle s_1 = s_2 = s$, we know therefore that the ONLY element of H in Stab(s) is e.