Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.5
Let $G$ be a group, $H$ a subgroup of $G$, and fix $g \in G$.
(1) Prove that $gHg^{-1} \leq G$ and that $|gHg^{-1}| = |H|$.
(2) Deduce that if $n \in \mathbb{Z}^+$ and $H$ is the unique subgroup of $G$ of order $n$ then $H$ is normal in $G$.
Solution:
(1) Note that $gHg^{-1}$ is not empty since $g1g^{-1} = 1 \in gHg^{-1}$. Now let $x = gag^{-1}$ and $y = gbg^{-1}$ be in $gHg^{-1}$, where $a,b \in H$. Then $$(gag^{-1})(gbg^{-1})^{-1} = g(ab^{-1})g^{-1} \in gHg^{-1},$$ and by the Subgroup Criterion we have $gHg^{-1} \leq G$.
We define a mapping $\varphi : H \rightarrow gHg^{-1}$ by $h \mapsto ghg^{-1}$. $\varphi$ is clearly surjective. Now suppose $\varphi(x) = \varphi(y)$; then $gxg^{-1} = gyg^{-1}$, so that $x = y$. Hence $\varphi$ is injective. Thus $|H| = |gHg^{-1}|$.
(2) Suppose now that $n \in \mathbb{Z}^+$ and that $G$ has a unique subgroup $H$ of order $n$. For all $g \in G$, $gHg^{-1}$ is a subgroup of order $n$, hence $gHg^{-1} = H$. Thus $H$ is normal in $G$.
