Solution
= Solution
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 cosets
$$
A,\qquad A(e,-1),
$$
and therefore has index two. Every index-two subgroup is normal, so
$$
\boxed{A\triangleleft D(A)}.
$$