If $G$ is a finite group and acting transitively on a set $X$ with $|X|>1$. then I have two question :-
- There is some element of $G$ in which fixes no element of $X$.
- Give a counter-example to this claim when $G$ and $X$ are infinite.
Any help will be appreciated.
This follows from Burnside's lemma. Since there is a single orbit, we have $$ \lvert G \rvert = \sum_{g \in G} X^g $$ where $X^g$ is the number of elements in $X$ fixed by $g$. Note $X^1 = \lvert X \rvert > 1$, and so if $X^g \geq 1$ for all $g \in G$, we have $$ \sum_{g \in G} X^g \geq \lvert X \rvert + \lvert G \rvert - 1 > \lvert G \rvert $$ which is a contradiction. Thus there is an element of $G$ that fixes no element.
For a counterexample: let $X = \mathbb{Z}$ and let $G$ be the group of all permutations of $\mathbb{Z}$ that only touch finitely many elements. Then all transpositions are in $G$, and so $G$ acts transitively on $\mathbb{Z}$, but each element of $G$ fixes infinitely many elements on $\mathbb{Z}$.