**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.7**

Let $G$ be a group and let $M,N \leq G$ be normal such that $G = MN$. Prove that $G/(M \cap N) \cong (G/M) \times (G/N)$.

Solution: Define $\varphi : G \rightarrow G/M \times G/N$ by $g \mapsto (gM,gN)$; certainly $\varphi$ is a homomorphism. If $g \in \mathsf{ker}\ \varphi$, then $(gM,gN) = (M,N)$, so that $g \in M \cap N$. Conversely, $M \cap N$ is contained in $\mathsf{ker}\ \varphi$.

Now we argue that $\varphi$ is surjective. To see this, let $(g_1M,g_2N) \in G/M \times G/N$. Since $G = MN$, and since $M$ and $N$ are normal, we have $$(g_1M,g_2N) = (m_1n_1M,m_2n_2N) = (n_1m_3M,m_2n_2N) = (n_1M,m_2N)$$ for some $m_i \in M$ and $n_i \in N$. Let $g = m_2n_1$; now $m_2n_1 = n_1m_4$ for some $m_4 \in M$, and evidently $\varphi(g) = (g_1M,g_2N)$. So $\varphi$ is surjective. By the first isomorphism theorem, we have $$G/(M \cap N) \cong G/M \times G/N.$$