Showing that a transitive abelian permutation group is necessarily regular

Question

I am trying to show that a transitive, abelian permutation group acting on a set $X$ is necessarily regular, given this hint: 'Given $g \in G$, consider the set $X^g:=\{x \in X\,|\,gx=x\}$. Show that if $G$ is transitive and abelian, then the only possibilities for $X^g$ are $\varnothing$ or $X$.'

I know that $X^g=X$ iff $g$ is the identity of $G$. And...that's about all I've got. I don't even see how, once the above is shown to be true, the result follows.

I would really appreciate some hints to help me in the right direction.

This is from Isaacs' Algebra: A Graduate course. He introduces permutation groups first, but group actions haven't entered the scene yet. I am alright with a hint in terms of actions though.

Thanks!

Answer 1

Hint: Let $1\not=g \in G$ and suppose $gx=x$ for some $x\in X$. Then take another element $y\in X$. Using the fact that $G$ is transitive, we have that some other element $h\in G$ such that $hy=x$. Now use the fact that $G$ is abelian to derive a contradiction.

Answer 2

Hint. You know $G$ is a permutation group, so only the identity fixes every point. Furthermore, you know that the intersection $$ \bigcap_{x \in X} G_{x} $$ of the stabilizers $G_{x} = \{ g \in G : g x = x \}$ is exactly the set of elements in $G$ that fix every point, so it is $\{ 1 \}$.

Spoiler

Now you know $G$ to be transitive on $X$. So once you fix $x_{0} \in X$, you will have $X = \{ g x_{0} : g \in G \}$. Thus $$\{1\} = \bigcap_{x \in X} G_{x} = \bigcap_{g \in G} G_{g x_{0}} = \bigcap_{g \in G} g G_{x_{0}} g^{-1} = \dots$$

Answer 3

Given $G \le Sym(X)$ is transitive and abelian, to show $G$ is regular, it suffices to show $G$ is semiregular, i.e. to show that $G_x=1$ for some $x \in X$. In other words, we need to show that if $g \in G$ fixes any element $x$, then it fixes all the remaining elements also, i.e. that $X^g=X$ or $=\phi$, $\forall g \in G$.

Suppose $g \in G$ fixes an element $x \in X$. So $x^g=x$. Let $y \in X$. We show $g$ fixes $y$ also. By transitivity of $G$, there is some $h \in G, h: x \mapsto y$. Hence $y^g = x^{hg}=x^{gh}=x^h=y$, where $hg=gh$ because $G$ is abelian. Thus, $g=1$ (i.e. $X^g=X$).

Answer 4

Complete set of hints:

1. What is the invariant subset $X^{g}=\{g\in G\mid g(x)=x\}$ like when…

   a. $g=e$?

   b. $g\ne e$?

2. For any fixed $x,y\in X$, there certainly exists $h\in G$ such that $h(x)=y$.
   What is $T=\{g\in G\mid g(x)=y\}$ and how many elements are there in $T$?

   a. Keep in mind that the goal is to show that the permutation group $G$ on $X$ is regular. What properties must $T$ have?

   b. To help this part of the proof, note the striking resemblance between $\{g\in G\mid g(x)=x\}$ and $\{g\in G\mid g(x)=y\}$.

Answer 5

Let $gx=y$, $hx=y$, where $g,h$-elements of the group $G$; $x,y$-elements of the set $A$. $G$ acts on $A$ Let $t$ some element of $A$. As $G$-transitive,there exist $f$ in $G$ that $t=fx$ As $G$ is abelian we have: $$gt=g(fx)=(gf)x=(fg)x=f(gx)=fy=f(hx)=(fh)x=(hf)x=h(fx)=ht$$ So for any $t$ in $A$ we have $gt=ht$, so $g=t$, then $G$-regular