\begindocument
\beginsolution Let $|G| = p^2$. The center $Z(G)$ is nontrivial by the class equation (since $|G| = |Z(G)| + \sum |G:C_G(g_i)|$, each term divisible by $p$). So $|Z(G)| = p$ or $p^2$.
\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 $D_8 = \langle r, s \mid r^4 = s^2 = 1, srs = r^-1 \rangle$. The center $Z(D_8)$ consists of elements commuting with all group elements.
Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$. \endsolution
\section*Chapter 4: Cyclic Groups and Properties of Subgroups \addcontentslinetocsectionChapter 4: Cyclic Groups
\beginsolution Recall: \beginitemize \item Centralizer: $C_G(H) = \ g \in G \mid gh = hg \ \forall h \in H \$. \item Normalizer: $N_G(H) = \ g \in G \mid gHg^-1 = H \$. \enditemize If $g \in C_G(H)$, then for all $h \in H$, $ghg^-1 = h \in H$, so $gHg^-1 = H$. Hence $g \in N_G(H)$. Therefore $C_G(H) \subseteq N_G(H)$. Both are subgroups of $G$, so $C_G(H) \le N_G(H)$. \endsolution
Check powers of $r$: $r$ does not commute with $s$ since $srs = r^-1 \ne r$ unless $r^2=1$, but $r^2$ has order 2. Compute $r^2 s = s r^-2 = s r^2$ (since $r^-2=r^2$), so $r^2$ commutes with $s$. Also $r^2$ commutes with $r$, thus with all elements. $r$ and $r^3$ are not central. $s$ is not central (doesn’t commute with $r$). Similarly $rs$ not central.
\begindocument
\beginsolution Let $|G| = p^2$. The center $Z(G)$ is nontrivial by the class equation (since $|G| = |Z(G)| + \sum |G:C_G(g_i)|$, each term divisible by $p$). So $|Z(G)| = p$ or $p^2$.
\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 $D_8 = \langle r, s \mid r^4 = s^2 = 1, srs = r^-1 \rangle$. The center $Z(D_8)$ consists of elements commuting with all group elements.
Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$. \endsolution
\section*Chapter 4: Cyclic Groups and Properties of Subgroups \addcontentslinetocsectionChapter 4: Cyclic Groups
\beginsolution Recall: \beginitemize \item Centralizer: $C_G(H) = \ g \in G \mid gh = hg \ \forall h \in H \$. \item Normalizer: $N_G(H) = \ g \in G \mid gHg^-1 = H \$. \enditemize If $g \in C_G(H)$, then for all $h \in H$, $ghg^-1 = h \in H$, so $gHg^-1 = H$. Hence $g \in N_G(H)$. Therefore $C_G(H) \subseteq N_G(H)$. Both are subgroups of $G$, so $C_G(H) \le N_G(H)$. \endsolution
Check powers of $r$: $r$ does not commute with $s$ since $srs = r^-1 \ne r$ unless $r^2=1$, but $r^2$ has order 2. Compute $r^2 s = s r^-2 = s r^2$ (since $r^-2=r^2$), so $r^2$ commutes with $s$. Also $r^2$ commutes with $r$, thus with all elements. $r$ and $r^3$ are not central. $s$ is not central (doesn’t commute with $r$). Similarly $rs$ not central.