Let G be a transitive subgroup of S_n where n>1. Show that G contains a permutation without fixpoints, i.e., where all cycles has length >1.
that's a quick result of Cauchy-Frobenius lemma (wrongly known as Burnside's lemma). let $\displaystyle X = \{1,2, ... , n \}.$ since $\displaystyle G$ is transitive, the number of orbits is 1. so we must have $\displaystyle |G|= \sum_{g \in G} |X^g|.$
let $\displaystyle \iota$ be the identity permutation. we have $\displaystyle X^{\iota}=|X|=n > 1.$ if for every $\displaystyle g \in G: \ |X^g| \geq 1,$ then $\displaystyle |G| > \sum_{g \in G} 1 = |G|.$ contradiction! so there exists $\displaystyle g \in G$ such that $\displaystyle |X^g|=0.$ Q.E.D.
Because it is transtive. Two elements are in the same orbit if and only if $\displaystyle y=gx$, since the group is transitive it means for any $\displaystyle x,y$ there is $\displaystyle g$ so that $\displaystyle y=gx$. Hence all elements are in the same obrit.
Because $\displaystyle |G| = \sum_{g\in G}|X^g|$ , $\displaystyle \sum_{g\in G} |X^g| > \sum_{g\in G}1$ (just see what happens for $\displaystyle g=e$). But $\displaystyle \sum_{g\in G}1 = |G|$ and so $\displaystyle |G| > |G|$.How does this follow?