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Let a finite group $G$ act transitively on a finite set $S$ with $|S| \geq 2$. The problem is to show that not every $g \in G$ can have a fixed point in this action. I proved this on my own, but I'm looking for alternate proofs.

My proof:

$G$ acts transitively so we can write $S = G.s$.

Assume that every $g \in G$ has a fixed point $h_g.s$. Then $g \in \operatorname{Stab}(h_g.s) = h_g \operatorname{Stab}(s) h_g^{-1}$. It turns out that $G$ is equal to the union of conjugates of $\operatorname{Stab}(s)$, but that can't happen unless $|S| = 1$.

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ante.ceperic
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  • We know that if the action is transitive, then the number of orbits is $1$, thus $1= \frac{1}{|G|}\sum_{g\in G} \pi(g)$ where $\pi$ is the corresponding permutation character. Then there must be some $g\in G$ with $\pi(g)=0$, so there exists an element which fix nothing. – Levent Jun 19 '16 at 19:08
  • Some other proofs here: http://math.stackexchange.com/questions/1820572/g-acts-on-x-transitively-then-there-exists-some-element-that-does-not-have-any – Ben West Jun 19 '16 at 19:08

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Let $t$ denote the number of orbits of $G$ acting on $S$. By the Cauchy-Frobenius Lemma (the lemma that is not Burnside's), $t = \frac{1}{|G|} \sum_{g \in G} |fix(g)|$, where $|fix(g)|$ is the number of points in $S$ that are fixed by $g$. In other words, the number of orbits is equal to the average value of the number of fixed points of permutations in $G$.

When the group action is transitive, $t=1$. The identity permutation $1_G$ fixes more than 1 point (assuming $|S| \ge 2$). Because the identity permutation contributes a value larger than 1, in order for the average value to be equal to 1, some permutation must contribute a value less than 1, ie, some permutation in $G$ has 0 fixed points.

Ashwin Ganesan
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