# [Math] How to prove that $MN/M \cap N \cong (MN/M) \times (MN/ N )$ where $M,N$ are normal subgroups of $G$

abstract-algebrafinite-groupsgroup-theoryproof-verification

I want to prove the following exercise from Dummit & Foote's Abstract Algebra:

Let $M$ and $N$ be normal subgroups of $G$ such that $G=MN$. Prove that
$G/M \cap N \cong (G/M) \times (G/N).$ [Draw the lattice]

I first trying following the hint of "drawing the lattice" which led me to apply the second (or diamond) isomorphism theorem twice as below:

Using the Second Isomorphism Theorem we have

\begin{equation}
G/N=MN/N \cong M/M \cap N
\end{equation}
as well as
\begin{equation}
G/M=NM/M \cong N/M \cap N
\end{equation}

so that is suffices to prove $G/M \cap N \cong N/M \cap N \times M \cong M \cap N$. However I couldn't get any further with this approach.

I decided to try something else, relying on a previous exercise:

Using Exercise 4 it suffices to prove
\begin{equation}
G/M \cap N \cong (G \times G)/(M \times N).
\end{equation}

We define $\varphi:G \to (G \times G)/(M \times N)$ by $\varphi(g)=(g,g)M \times N$. It is a homomorphism because $\varphi(gh)=(gh,gh)M \times N=(g,g)(h,h)M \times N=\left[(g,g) M \times N \right] \left[ (h,h) M \times N \right]=\varphi(g)\varphi(h)$. It is surjective obviously, because any coset $(g,g) M \times N$ is the image of $g$. The kernel is all $g \in G$ such that $(g,g) \in M \times N$ which is simply the intersection $M \cap N$. The First Isomorphism theorem gives the result.

I have two questions regarding the above:

1. Is it possible to get a complete proof following the Authors' Hint?
2. Is my alternative proof correct? If so, please help me correct it.

Thanks!

For surjectivity, consider any element of $(xM,yN)\in(G/M)\times(G/N)$. Since $G=MN$ there exist $m\in M$, $n\in N$ such that $m^{-1}n=xy^{-1}$. Let $z=mx=ny$ and $$\phi(z(M\cap N))=(zM,zN)$$ but $zM=mxM=xM$, since $M$ is normal, and similarly, $zN=nyN=yN$, hence $\phi$ is surjective.