## 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.

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

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

• nbloomf  On October 13, 2011 at 10:18 am

Thanks!

• 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!