Exercise 4.3.24 ($G$ is not the union of the conjugates of any proper subgroup)

Proof. Let $H$ be a proper subgroup of the finite group $G$.

Assume for the sake of contradiction that $G\ne {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{g\in G}\mathit{𝑔𝐻}{g}^{-1}$. Since $G$ is finite, there is a maximal subgroup $M$ such that $H\le M$. Then $G={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{g\in G}\mathit{𝑔𝐻}{g}^{-1}\subseteq {\bigcup }_{g\in G}\mathit{𝑔𝑀}{g}^{-1}\subseteq G$, thus

If $M⊴G$, then $M={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{g\in G}\mathit{𝑔𝑀}{g}^{-1}=G$: this is impossible, since $M$ is maximal in $G$, which implies $M\ne G$. So $M$ is not normal in $G$. By Exercise 23,

Therefore

But $M$ is a proper subgroup of $G$, so

thus

or equivalenty

so

The contradiction $|G|<|G|$ shows that $G$ is not the union of the conjugates of any proper subgroup. □