Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 6 / a

A subgroup $N$ of $G$ is a normal subgroup if conjugation by every element preserves it:Equivalently, $gN=Ng$ for every $g\in G$. Here equality means equality of sets; it does not require each individual element of $N$ to commute with every element of $G$.

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 $x$ andThus $a{b}^{-1}\in Z(G)$, proving closure under the subgroup criterion. Moreover, if $z\in Z(G)$, then $xz{x}^{-1}=z$ for every $x\in G$. Hence conjugation fixes the center elementwise, soTherefore the center of a group is a normal subgroup.

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus