Every nontrivial transitive permutation group contains elements which are fixed point free

Let G be a transitive permutation group on the finite set A with |A| \geq 2. Show there is some \sigma \in G such that \sigma(a) \neq a for all a \in A. (Such an element is called fixed point free.)

First we prove a lemma.

Lemma: \sigma \mathsf{stab}(a) \sigma^{-1} = \mathsf{stab}(\sigma(a)). Proof: (\subseteq) if \tau \in \mathsf{stab}(a), note that (\sigma \tau \sigma^{-1})(\sigma(a)) = \sigma(\tau(a)) = \sigma(a). Thus \sigma \tau \sigma^{-1} \in \mathsf{stab}(\sigma(a)). (\supseteq) Let \tau \in \mathsf{stab}(\sigma(a)). Then \tau(\sigma(a)) = \sigma(a), so that (\sigma^{-1} \tau \sigma)(a) = a; hence \sigma^{-1} \tau \sigma \in \mathsf{stab}(a), so that \tau \in \sigma \mathsf{stab}(a) \sigma^{-1}. \square

Suppose by way of contradiction that for all \sigma \in G there exists an element a \in A such that \sigma(a) = a. Then G = \bigcup_{a \in A} \mathsf{stab}(a).

Now because this action is transitive, if we fix b \in A, then as \sigma ranges over G, \sigma \cdot b is arbitrary in A. So in fact, G = \bigcup_{\sigma \in G} \mathsf{stab}(\sigma(b)) = \bigcup_{\sigma \in G} \sigma \mathsf{stab}(b) \sigma^{-1}. Now because the action is transitive and |A| \geq 2, we know that \mathsf{stab}(b) is a proper subgroup. Finally, G \leq S_A is finite. By a previous theorem, we have a contradiction. Thus there exists an element \sigma \in G that is fixed point free.

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  • Peter  On October 12, 2011 at 11:25 pm

    In your last equation with the unions, there is an extra stab.

  • Gobi Ree  On December 12, 2011 at 1:34 am

    A typo: \bigcup_{g\in G} should be \bigcup_{\sigma \in G}.

    • nbloomf  On December 12, 2011 at 11:38 am

      Thank you!

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