= Solution
\b[Conjugacy classes.] For $a\in G$, its <conjugacy class> is
$$
\boxed{\operatorname{Cl}(a)=\{gag^{-1}:g\in G\}.}
$$
Every element belongs to its own <conjugacy class>, by taking $g=e$. To prove uniqueness, suppose $x=gag^{-1}=hbh^{-1}$ belongs to both $\operatorname{Cl}(a)$ and $\operatorname{Cl}(b)$. Then
$$
b=(h^{-1}g)a(h^{-1}g)^{-1}.
$$
For every $t\in G$, this makes $tbt^{-1}$ a conjugate of $a$ as well. Hence $\operatorname{Cl}(b)\subseteq\operatorname{Cl}(a)$, and reversing the argument gives the other inclusion. Two <conjugacy classes> that meet are therefore equal, proving that the classes partition $G$.
\b[Abelian case.] If $G$ is an <abelian group>, then $gag^{-1}=agg^{-1}=a$ for every $g$, so $\boxed{\operatorname{Cl}(a)=\{a\}}$ for each $a$.
\b[Centre.] An element has a singleton <conjugacy class> precisely when $gzg^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual <centre of a group>:
$$
Z(G)=\{z\in G:gz=zg\text{ for every }g\in G\}.
$$
The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=gz^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a <subgroup>. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.
\b[Square symmetry.] Use the paper's notation $D_4$ for the eight-element <dihedral group> of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^js$ commutes with $r$, because
$$
(r^js)r=r^{j-1}s,\qquad r(r^js)=r^{j+1}s,
$$
and $r^{j-1}\ne r^{j+1}$. Therefore
$$
\boxed{Z(D_4)=\{e,r^2\}=\{\text{identity},\text{half-turn}\}.}
$$
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