Let $M$ and $N$ be two normal subgroups of $G$. Show that $M \cap N$ is also normal in $G$. Then show that $G/(M \cap N) \cong (G/M) \times (G/N)$

Question

Let $M$ and $N$ be two normal subgroups of $G$. Show that $M \cap N$ is also normal in $G$. Furthermore,if $G=MN$ then show that $$G/(M \cap N) \cong (G/M) \times (G/N)$$ I manage to prove the first part. But for the two quotient groups to be isomorphic, I have no idea. Anyone can help ?

Answer 1

The second assertion is true if we add one more condition that $G=MN$, if we define the map $$f:G\to (G/M) \times (G/N)$$ such that $$f(g)=(gM,gN)$$ then $f$ is a homomorphism and it is easily seen that $kerf=M\cap N$, now if we can prove that $f$ is onto we can use the first isomorphism theorem to get the result.

will let $(xM,yN)\in (G/M) \times (G/N)$ now since $G=MN$ we have $x=mn$ and $y=m'n'$ but then we have$$xM=mnM=n(n^{-1}mn)M=nM$$ since $m\in M$ and $M$ is normal,similarly we get$$yN=m'N$$ now consider $$f(m'n)=(m'nM,m'nN)=(n(n^{-1}m'n)M,m'nN)=(nM,m'N)=(xM,yN)$$ again since $m\in M$ and $M$ is normal and hence that map is onto and we have $$G/(M \cap N) \cong (G/M) \times (G/N).$$

Answer 2

Second assertion is not true. E.g., for $M=N=1$ you get from your isomorphism $G\cong G\times G$.

Answer 3

Part two is false: $6\Bbb Z\cap 3\Bbb Z=6 \Bbb Z$, but $\Bbb Z /6\Bbb Z$ is not isomorphic to $\Bbb Z /3\Bbb Z\times \Bbb Z /6\Bbb Z$

The former has 6 elements, but the latter has 18.

Answer 4

Can you think of a homomorphism from the left side to the right side? In particular if $[g]$ represents the equivalence class of some $g \in G$ on the left side, where should it be sent to on the right side? Can you prove that this is well-defined? An injection? A surjection?

Answer 5

Probably the task is to show that $M \cap N$ is normal and that $G/(M \cap N)$ embeds into $G/M \times G/N$ (do you really have reproduced the exercise correctly?). In fact, both can be done in one step: Consider the diagonal map $G \to G/M \times G/N$, $g \mapsto ([g],[g])$. It is a homomorphism, with kernel $M \cap N$. Done.