Let $G$ be a finite group.
1.Suppose that $G$ acts transitively on a finite set $X$ with $|X|\geq2$.Show that there is an element $g\in G$ with no fixed points on $X$.
2.Let $H$ be a proper subgroup of $G$. Prove that
$G \neq$ $\cup_{x\in G} xHx^{-1}$
I have done the first part using Burnside Lemma. I am unable to do the second. The book asks to use the first part. I have no idea how to do that. Thanks!!!
If $H$ is normal the result is trivial because $H$ exhausts the set of all conjugates, and $H$ is not the whole group. Otherwise note that $G$ acts transitively on the set of all conjugates $$\{xHx^{-1}|x\in G\}$$ The action is given by conjugation. An element $g$ with no fixed points for this action does not stabilize any of these conjugates, hence it cannot be contained in any of them. This is exactly what we wanted to prove (that such an element exists).