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As the title states, the question is the following: Let (G, X) be a transitive permutation group, where G is abelian. Show that (G, X) is "sharply regular".

First of all I want to notice that in my course we never saw what "sharply regular" means in relation to permutation groups. Wikipedia however, tells me that my professor perhaps meant "sharply transitive", so I'm going to solve the question assuming he meant this. Please tell me if it's possible that he meant something else.

Proof: So, I need to prove that: $\forall x, y \in X, \exists! g \in G: x^g = y$

Given that (G, X) is a transitive permutation group, we know already that there exists at least one such g. Now we have to show that there exists just one such g.

Suppose that there exists $h \in G: x^h = y$ as well. Then $x^{gh^{-1}} = y^{h^{-1}} = x$. Therefore it is clear that $gh^{-1} \in kernel$. (G, X) is a permutation group, which means that the kernel is trivial. Therefore: $gh^{-1} = e$. Now it is clear that g = h and the question is solved.

However, in this proof I did not use the fact that G is abelian. Something tells me that they gave me this information on purpose. So I assume I have made a mistake somewhere in my proof. Could anyone point out the mistake and perhaps show me how to proof this using the fact that G is commutative?

Uomo universale
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1 Answers1

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Let $x \in X$, and $G_{x}$ be the stabilizer of $x$. Since $G$ acts transitively, for any $y \in X$, there will be $g \in G$ such that $x^{g} = y$. Thus $G_{y} = G_{x^{g}} = (G_{x})^{g} = G_{x}$, as $G$ is abelian. Hence $G_{x}$ fixes all elements of $X$, so that $G_{x}$ is trivial.

Finish it by noting that if $y = x^{g}$, then the set of elements $h \in G$ such that $y = x^{h}$ is the coset $G_{x} g$.

Andreas Caranti
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