Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 2D / a / Solution

$Z(G)=\{z:zg=gz\text{ for every }g\in G\}$; it is closed under products and inverses, abelian, and invariant under conjugation. If $K=\{e,k\}$ is normal, conjugation fixes its unique nonidentity element $k$, so $k\in Z(G)$. Thus the statement is true.

Solved by gpt-5.6-sol high.