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
G/N=MN/N \cong M/M \cap N
as well as
G/M=NM/M \cong N/M \cap N
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
G/M \cap N \cong (G \times G)/(M \times N).
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:
- Is it possible to get a complete proof following the Authors' Hint?
- Is my alternative proof correct? If so, please help me correct it.