If $M/G_1$ and $M/G_2$ are Noetherians, then $M/(G_1\cap G_2)$ is Noetherian. I'm trying to prove if $M$ is a module, $G_1$, $G_2$ submodules of $M$  and $M/G_1$ and $M/G_2$ Noetherians, then $M/(G_1\cap G_2)$ is Noetherian.
I've tried by brute force (writing down explicitly an ascending chain), by second fundamental theorem of isomorphisms, etc... without success.
I really need help.
Thanks a lot.
 A: Consider the canonical map
$$\Phi \colon M \to M/G_1 \times M/G_2; \quad \Phi(m) = ([m]_{G_1},\, [m]_{G_2}).$$
The kernel is $\ker \Phi = G_1 \cap G_2$, so $\Phi$ induces an embedding
$$\varphi \colon M/(G_1 \cap G_2) \hookrightarrow M/G_1 \times M/G_2.$$

 $P = M/G_1 \times M/G_2$ is Noetherian (why?), $M/(G_1\cap G_2)$ is isomorphic to a submodule of $P$. A submodule of a Noetherian module is Noetherian (why?).

A: Consider the modules $M/(G_1 \cap G_2)$ and note that $G_1/(G_1 \cap G_2)$ is a submodule. Then we have
$$\frac{M/(G_1\cap G_2)}{G_1/(G_1\cap G_2)} \cong M/G_1$$
is Notehrian.
On the other hand, the third isomorphism theorem (for modules) gives us that $G_1/(G_1\cap G_2)$ is isomorphic to $(G_1+G_2)/G_2,$ which is a submodule of the Noetherian module $M/G_2$ and hence, is Noetherian. Then we deduce that $G_1/(G_1 \cap G_2)$ is Noetherian.
Finally, consider the exact sequence
$$0\rightarrow G_1/(G_1 \cap G_2) \xrightarrow{\iota} M/(G_1 \cap G_2)\xrightarrow{\rho} \frac{M/(G_1\cap G_2)}{G_1/(G_1\cap G_2)} \to 0,$$
where $\iota$ is the inclusion map and $\rho$ is the canonical projection.
Since the left and right modules of the exact sequences are both Noetherian, we deduce that de desired module is Noetherian.
