## The quotient of a product by a product is isomorphic to the product of quotients

Let $A$ and $B$ be groups, with $C \leq A$ and $D \leq B$ normal. Prove that $(C \times D) \leq (A \times B)$ is normal and that $(A \times B)/(C \times D) \cong (A/C) \times (B/D)$.

We have already seen that $C \times D$ is a subgroup. Now let $(a,b) \in A \times B$. we have $(a,b)(C \times D) = aC \times bD = Ca \times Db = (C \times D)(a,b)$, hence $C \times D$ is normal.

Define $\varphi : A \times B \rightarrow (A/C) \times (B/D)$ by $(a,b) \mapsto (aC,bD)$. Clearly $\varphi$ is surjective. We now show that $\mathsf{ker}\ \varphi = C \times D$. If $(a,b) \in \mathsf{ker}\ \varphi$, we have $(aC,bD) = (C,D)$, hence $a \in C$ and $b \in D$. Thus $(a,b) \in C \times D$. If $(a,b) \in C \times D$, then $(aC,bD) = (C,D)$. Thus $\mathsf{ker}\ \varphi = C \times D$, and by the First Isomorphism Theorem, $(A \times B)/(C \times D) \cong (A/C) \times (B/D)$.

I can’t recall which problem this was in, but it’s not so hard to see using the subgroup criterion: $C \times D$ contains $(1,1)$ and thus is not empty, and if $(a_1,b_1),(a_2,b_2) \in C \times D$, then $(a_1,b_1)(a_2,b_2)^{-1} = (a_1a_2^{-1},b_1b_2^{-1}) \in C \times D$.