TITLE: Union of the conjugates of a proper subgroup
QUESTION [51 upvotes]: Let G be a finite group and H be a proper subgroup. Prove that the union of the conjugates of H is not the whole of G.
Thanks for any help

REPLY [3 votes]: Lemma: the number of distinct conjugates of $H$ is equal to $|G:N_G(H)|$.
Pf of lemma: 
$\phi: G/N_G(H)\to \{gHg^{-1}\}$ with $\phi(gN_G(H))=gHg^{-1}$ is a bijection for the following reason:
$gN_G(H)=g'N_G(H)\iff \begin{cases}g'^{-1}g\in N_G(H)\\ g'g^{-1}\in N_G(H)\end{cases}\iff \begin{cases} g'^{-1}ghg^{-1}g'\in H&\forall h\in H\\ g'g^{-1} h'gg'^{-1}\in H&\forall h'\in H\end{cases}\iff $$\begin{cases} ghg^{-1}\in g'Hg'^{-1} &\forall h\in H\\g'h'g'^{-1}\in gHg^{-1}&\forall h'\in H \end{cases}\iff \begin{cases} gHg^{-1}\subset g'Hg'^{-1}\\ g'Hg'^{-1}\subset gHg^{-1} \end{cases} \iff gHg^{-1}= g'Hg'^{-1}$
So $\phi$ is well-defined and injective.
it's surjective for obvious reason.
Hence the number of distinct conjugates of $H$ is equal to $|G:N_G(H)|$.
$|\bigcup_{g\in G} gHg^{-1}|\leqslant \sum_{g\in G/N_G(H)} |gHg^{-1}|-|\bigcap_{g\in G} gHg^{-1}|= |G:N_G(H)| |H|-|\bigcap_{g\in G} gHg^{-1}|<\dfrac{|G|}{|N_G(H)|}|H|\leqslant \dfrac{|G|}{|H|}|H|= |G|$
Notice that
$\text{(1)}\forall g\in G/N_G(H), |gHg^{-1}|=|H| \text{ for obvious reasons}\\
\text{(2)} e\in \bigcap_{g\in G} gHg^{-1}\\
\text{(3)} H\leqslant N_G(H)$