**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.4**

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 $$(A \times B)/(C \times D) \cong (A/C) \times (B/D).$$

Solution: 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, $$(A \times B)/(C \times D) \cong (A/C) \times (B/D).$$