# Showing that a transitive abelian permutation group is necessarily regular

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!

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.

• Does this work? Suppose $1 \ne g \in G$, and $gx=x$ for some $x \in X$. Then for some $x \ne y \in X$, there exists another element $h \in G$ such that $hy=x$. But then $gx=x \Rightarrow gx=hy \Rightarrow h^{-1}gx=y \Rightarrow gh^{-1}x=y \Rightarrow gy=y$. But $x$ and $y$ are arbitrary, so $g$ must be the identity, contradicting our assumption that $g \ne 1$. So for $g \ne 1$ we must have $X^g=\varnothing$. – Alex Petzke Mar 8 '13 at 19:59
• @AlexPetzke You got it! – Alexander Gruber Mar 8 '13 at 20:42
• Nice, thanks for the hint. – Alex Petzke Mar 9 '13 at 2:40

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 $$\{1\} = \bigcap_{x \in X} G_{x} = \bigcap_{g \in G} G_{g x_{0}} = \bigcap_{g \in G} g G_{x_{0}} g^{-1} = \dots$$

• I'm not seeing where this is going (the spoiler wasn't much of a spoiler for me)...could you explain? – Alex Petzke Mar 10 '13 at 2:42
• @AlexPetzke, the stabilizers are all conjugate (since the group acts transitively), so they are the same in an abelian group. Thus the stabilizer of a single point fixes all points, and thus must be $\{ 1\}$, so that the group acts regularly. – Andreas Caranti Mar 10 '13 at 7:28
• I'm not used to all the details of group actions yet, but with some help from Wikipedia, I get it now. Thanks. – Alex Petzke Mar 11 '13 at 2:18

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. $X^g=X$).

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 $\{g\in G\mid g(x)=y\}$.