Normal subgroup fixes every element of a set.

Question

Let $G$ be a group and $H \unlhd G$. Suppose that $G$ acts transitively on a set $\Omega$ and that $H$ fixes some $x\in \Omega$, where $h\cdot x=x$ for all $h\in H.$

Prove that $H$ fixes every element of $\Omega$.

I know that for any $y\in\Omega$ since $G$ is transitive, there is some $g\in G$ such that $g\cdot y=x$, where $x$ is the point which $H$ fixes.

Answer 1

Here are some ideas:

1. To say that $H$ fixes $x$ means that $H$ is contained in the stabiliser $G_x$ of $x$ in $G$.

2. To say that $H$ fixes every element of $\Omega$ means that $H$ is contained in the kernel of the action of $G$ on $\Omega$.

3. Since $G$ is transitive, the kernel of the action of $G$ on $\Omega$ can be expressed as a certain intersection of point stabilisers contained in $G_x$. Remember also that $H$ is assumed to be normal in $G$.

Can you take it from here?

Answer 2

Suppose $H$ fixes $x$. Let $h \in H$ and $y \in \Omega$. We need to show that $h$ fixes $y$. As the action is transitive, there is some $g \in G$ such that $g \cdot x = y$. Let $h' = g^{-1}hg$. Then $h' \in H$ because $H$ is normal. Therefore we have $h' \cdot x = x$. Now observe that $$ h \cdot y = (gh'g^{-1}) \cdot y = (gh'g^{-1}) \cdot (g\cdot x) = (gh') \cdot x = g \cdot (h' \cdot x) = g \cdot x = y $$ as required.

Answer 3

Here’s a hint.

You know that $h\cdot x = x$ for any $h\in H$. You also know that any $y\in\Omega$ can be expressed as $g\cdot x$ for some $g\in G$.

Using both of these facts, play around with the expression $h\cdot y = y$. If you get stuck, here’s the idea:

If we unwind the facts given, the expression $h\cdot y = y$ looks like $$h\cdot (g\cdot x) = g\cdot x$$ which can be rewritten as $$(ghg^{-1})\cdot x = x.$$ Thus, given some $h\in H$ and some $y\in\Omega$ where $y = g\cdot x$, to prove that $h\cdot y = y$, we first see that $h = gh’g^{-1}$ for some $h’\in H$. Thus, we have $$h\cdot y = gh’g^{-1}\cdot y = (gh’)\cdot(g^{-1}\cdot y) = gh’\cdot x = g\cdot x = y$$ which shows that any element of $H$ fixes arbitrary elements of $\Omega$.