= Solution
For $\epsilon=-1$,
$$
(a,-1)^2=(aa^{-1},1)=(e,1),
$$
so every element outside $A$ has order two.
The element $(e,-1)$ commutes with $(a,1)$ exactly when
$$
(a,-1)=(a^{-1},-1),
$$
or $a^2=e$. 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
$$
\boxed{
Z(D(A))=\{(a,1):a^2=e\}}.
$$
Back to article page
