\beginsolution Let $|H| = n$ and suppose $H$ is the only subgroup of $G$ with order $n$. For any $g \in G$, consider $gHg^-1$. Conjugation is an automorphism of $G$, so $|gHg^-1| = |H| = n$. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$. By uniqueness, $gHg^-1 = H$ for all $g \in G$. Hence $H \trianglelefteq G$. \endsolution
\newpage \section*Supplementary Problems for Chapter 4 Dummit And Foote Solutions Chapter 4 Overleaf High Quality
\subsection*Exercise 4.3.12 \textitProve that if $H$ is the unique subgroup of a finite group $G$ of order $n$, then $H$ is normal in $G$. \beginsolution Let $|H| = n$ and suppose $H$
\subsection*Exercise 4.1.1 \textitProve that every cyclic group is abelian. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$
% Solution environment \newtcolorboxsolution colback=gray!5, colframe=blue!30!black, arc=2mm, title=Solution, fonttitle=\bfseries
If $|Z(G)| = p^2$, then $G$ is abelian. If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, hence is cyclic. A well-known lemma states: if $G/Z(G)$ is cyclic, then $G$ is abelian. So $G$ is abelian in both cases. \endsolution