Union of the Conjugates of a Proper Subgroup

Proposition

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 $G$.

\begin{proof} In the finite case, let $|H|=k$; then $|G|=kn$. There are at most $n$ distinct conjugates. Since the identity element is in all of the conjugates, the union of the conjugates of $H$ has at most

$$n(k-1)+1=nk-n+1\text{ element }$$

and since we are assuming $n>1$, it follows that

$$\left|\bigcup _{g\in G}gH{g}^{-1}\right|\le nk-(n-1)<nk=|G|$$

so the union cannot equal all of $G$. \end{proof}

Remark. If $|G|$ is infinite, we have the following counterexample: $G=\mathrm{Sym}(\mathbb{Z})$ and $H=⟨(12)⟩$.