Exercise 3.2.5(If $H$ is the unique subgroup of $G$ of order $n$ then $H \unlhd G$)

Proof. Let $H$ be a subgroup of $G$ and fix some element $g\in G$.

(a)

The inner automorphism ${\gamma }_g:G\to G$ defined by ${\gamma }_g(x)=\mathit{𝑔𝑥}{g}^{-1}$ maps $H$ onto $\mathit{𝑔𝐻}{g}^{-1}$. Since ${\gamma }_g$ is an automorphism, $\mathit{𝑔𝐻}{g}^{-1}={\gamma }_g(H)$ is a subgroup of $G$ of the same order as $H$.

(b)

We suppose that $H$ is the unique subgroup of $G$ of order $n$. Then for any $g\in G$, by part (a), $\mathit{𝑔𝐻}{g}^{-1}={\gamma }_g(H)$ is a subgroup of $G$ of order $n$. Hence $\mathit{𝑔𝐻}{g}^{-1}=H$. This proves that $H⊴G$.