Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 8D / a

Identify $A$ with $\{(a,1):a\in A\}$. Componentwise use of the stated multiplication shows that this set contains the identity and is closed under products and inverses, so it is a subgroup. It has the two cosetsand therefore has index two. Every index-two subgroup is normal, so

For $ϵ=-1$,so every element outside $A$ has order two.

The element $(e,-1)$ commutes with $(a,1)$ exactly whenor $a^2=e$. Since $A$ itself is abelian, $D(A)$ is abelian exactly when every $a\in A$ satisfies $a^2=e$.

Suppose $D(A)$ is nonabelian. An element $(a,1)$ is central exactly when $a=a^{-1}$, equivalently $a^2=e$. No element $(a,-1)$ is central, because commuting with every $(b,1)$ would require $b^2=e$ for every $b$, which would make the whole group abelian. Hence