If $\displaystyle G $ is a finite group and $\displaystyle H$ is a subgroup of $\displaystyle G$ then prove that $\displaystyle \displaystyle G \neq \bigcup\limits_{a \in G} aHa^{-1}$
Of course, we assume H is a PROPER subgroup.
Let G act by conjugation on the set X of all its proper subgroups; then we get that
$\displaystyle s:=|Orb(H)|=[G:N_G(H)]\leq [G:H]=r$ , say, and since $\displaystyle |aHa^{-1}|=|H|\,\,\,\forall\,a\in G$ ,we get that:
$\displaystyle |\bigcup aHa^{-1}|\leq 1+s(|H|-1)\leq 1+r(|H|-1)=1+r|H|-r=|G|-(r-1)<|G|$
(Question: why $\displaystyle |\bigcup aHa^{-1}|\leq 1+s(|H|-1)$ ??)
Tonio