1. ## Transitive subgroup

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.

2. Originally Posted by peteryellow

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 $X = \{1,2, ... , n \}.$ since $G$ is transitive, the number of orbits is 1. so we must have $|G|= \sum_{g \in G} |X^g|.$

let $\iota$ be the identity permutation. we have $X^{\iota}=|X|=n > 1.$ if for every $g \in G: \ |X^g| \geq 1,$ then $|G| > \sum_{g \in G} 1 = |G|.$ contradiction! so there exists $g \in G$ such that $|X^g|=0.$ Q.E.D.

3. Originally Posted by NonCommAlg
since $G$ is transitive, the number of orbits is 1.
How do you have thet number of orbit is 1.

Originally Posted by NonCommAlg
$|G| > \sum_{g \in G} 1 = |G|.$ contradiction!
How does this follow?

4. Originally Posted by peteryellow
How do you have thet number of orbit is 1.
Because it is transtive. Two elements are in the same orbit if and only if $y=gx$, since the group is transitive it means for any $x,y$ there is $g$ so that $y=gx$. Hence all elements are in the same obrit.

How does this follow?
Because $|G| = \sum_{g\in G}|X^g|$ , $\sum_{g\in G} |X^g| > \sum_{g\in G}1$ (just see what happens for $g=e$). But $\sum_{g\in G}1 = |G|$ and so $|G| > |G|$.

5. Thanks NoncommutativeAlgebra and PerfectHacker.