= Solution
{c}
The specified <set> $A$ is the <center of a group> $Z(G)$. It contains $e$. If $a,b\in A$ and $y\in G$, then
$$
(ab)y=a(by)=a(yb)=(ay)b=y(ab),
$$
so $ab\in A$. From $ay=ya$, multiplying on the left and right by $a^{-1}$ gives $ya^{-1}=a^{-1}y$, so $a^{-1}\in A$. Hence $A$ is a <subgroup>.
For $g\in G$ and $a\in A$, the defining <commutativity> gives $gag^{-1}=agg^{-1}=a$. Thus the image of each element of $A$ under <conjugation> is in $A$, and indeed $gAg^{-1}=A$. Consequently \b[$A=Z(G)$ is a <normal subgroup> of $G$].
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