= Solution
{c}
The <center of a group> consists of its elements commuting with every element. First check it is a <subgroup>. The <identity element> is central. If $a,b\in Z(G)$ and $x\in G$, then $b^{-1}$ commutes with $x$ and
$$
(ab^{-1})x=a(xb^{-1})=x(ab^{-1}).
$$
Thus $ab^{-1}\in Z(G)$, proving closure under the subgroup criterion. Moreover, if $z\in Z(G)$, then $xzx^{-1}=z$ for every $x\in G$. Hence conjugation fixes the center elementwise, so
$$
\boxed{xZ(G)x^{-1}=Z(G)\quad(x\in G).}
$$
Therefore the <center of a group> is a <normal subgroup>.
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