Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$. \endsolution
\subsection*Exercise 4.1.3 \textitFind all subgroups of $\Z_12$ and draw the subgroup lattice. Dummit And Foote Solutions Chapter 4 Overleaf High Quality
\begindocument
\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 $G$. Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$
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. Check powers of $r$: $r$ does not commute
\subsection*Exercise 4.8.3 \textitShow that $\Inn(G) \cong G/Z(G)$.
\beginsolution $\Z_12 = \0,1,2,\dots,11\$ under addition modulo 12. By the fundamental theorem of cyclic groups, for each positive divisor $d$ of 12, there is exactly one subgroup of order $d$, namely $\langle 12/d \rangle$.