A group is not the union of conjugates of any proper subgroup

Assume $H$ is a proper subgroup of the finite group $G$. Prove that $G \neq \bigcup_{g \in G} gHg^{-1}$; i.e., $G$ is not the union of the conjugates of any proper subgroup.

There exists a maximal subgroup $M$ containing $H$. If $M$ is normal in $G$, then $\bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1} = M \neq G$. If $M$ is not normal, we still have $\bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1}$. By the previous exercise, $\bigcup_{g \in G} gMg^{-1}$ contains at most $(|M|-1)[G : M]$ nonidentity elements. Thus $|\bigcup_{g \in G} gHg^{-1}| \leq |G| - [G : M] + 1 < |G|$, since $[G:M] \geq 2$.

In particular, because $G$ is finite, $G \neq \bigcup_{g \in G} gHg^{-1}$. Thus $G$ is not the union of all conjugates of any proper subgroup.