If you find any mistakes, please make a comment! Thank you.

If a group has a unique subgroup of a given order, then that subgroup is normal

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 $G$.


(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 $G$.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. How is latex supported here? I have an other answer for (a) that i'd like sharing. Thanks

Leave a Reply

Close Menu