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I want to prove the following result:

$G$ is a finite, abelian group acting faithfully on a finite set $X$. If the action is transitive then show $|G| = |X|$.

A proof of this result has been given here, but I have attempted this question in an alternative way, and something must be wrong since I have not used all the parts of the question, namely that $G$ is abelian. However, I can't see what is wrong with the proof.

Attempt at proof

Since $G$ is transitive, $\forall x, y \in X, \exists g \in G$ such that $g(x) = y$.

So, since $G$ and $X$ are finite, we must require that $|G| \geq |X|$, as illustrated:

G bigger than or the same size as X

Now suppose $|G| > |X|$. Then this means $\forall x \in X$ (specifically with $x$ not the identity), $ \exists$ $ g, h \in G$ such that $g(x) = h(x)$ by pigeonhole principle, with $g \neq h$.

So $h^{-1}g(x) = x$, so $h^{-1}g = e$ since $G$ acts faithfully, and we have assumed $x$ is not the identity. But then $g = h$, which is a contradiction.

Therefore $|G| = |X|$.

I don't think this is right, since I have not used $G$ is abelian, but I can't see what is wrong with this. Where is the error?

Robin
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    You don't need $G$ to be abelian for the result to be true. – Stephen Nov 19 '21 at 15:06
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    An illustration is not a proof. In addition, you have the definition of "faithful" incorrect. The action is faithful if and only if $gx=x$ for all $x\in X$ implies $g=e$. You only have $h^{-1}gx=x$ for a single $x$. The condition you are using is called "fixed-point-free", not "faithful". – Arturo Magidin Nov 19 '21 at 15:18
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    The comment by @Steven is wrong. The statement is not true if the group is not abelian, take e.g. $S_3$ acting on the points ${1,2,3}$. – ahulpke Nov 19 '21 at 15:19

1 Answers1

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Your error is in the conclusion of $h^{-1}(g(x))=x$ implying that $h^{-1}g=e$. Take for example $g=(1,2)$ and $h=(1,2,3)$.

More abstractly, the problem is in quantifying $h$. You have $\forall g\forall x\exists h$,that is $h$ depends on $x$, and thus you cannot deduce that from a statement for one particular $x$ it would follow for all.

ahulpke
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