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Spring 2016 - Problem 9

group theory

Show that if $G$ is a finite group acting transitively on a set $X$ with at least two elements, then there exists $g \in G$ which fixes no point of $X$ .

Solution.

Since $G$ is transitive, there is only one orbit. Thus, by Burnside's lemma,

1 = \frac{1}{\abs{G}} \sum_{g \in G} \abs{X^g} = \frac{1}{\abs{G}} \p{\abs{X} + \sum_{g \in G \setminus \set{e}} \abs{X^g}}

where $X^g$ is the set of fixed points of $g$ . If the claim were false, then $\abs{X^g} \geq 1$ for each $g \in G$ , but this means

\abs{G} \geq \abs{X} + \p{\abs{G} - 1} \geq 2 + \abs{G} - 1 = \abs{G} + 1,

which is a contradiction. Hence, the claim holds.