Two normal subgroups and isomorphism theorem Question

Let $N_1$ and $N_2$ be normal subgroups of $G$.
Prove that $N_1N_2/(N_1\cap N_2) \cong (N_1N_2/N_1)\oplus (N_1N_2/N_2)$.

I think the homomorphism must be $\phi : N_1N_2 \to (N_1N_2/N_1)\oplus (N_1N_2/N_2)$
such that $\phi(n_1n_2)=(n_1n_2N_1,n_1n_2N_2)$
Then by 1st isomorphism theorem, $N_1N_2/ker\phi \cong Im\phi$
and $ker\phi = N_1 \cap N_2$
But it is hard to show that $\phi$ is surjective.
What should I do?
 A: As you note, it is a simple application of the first isomorphism theorem.
You have correctly defined a morphism
$$
\varphi : N_{1} N_{2} \to \dfrac{N_1 N_2}{N_1} \times \dfrac{N_1 N_2}{N_2}
$$
by
$$
\varphi(g) = (g N_{1}, g N_{2}),
$$
and checked that the kernel is $N_{1} \cap N_{2}$. 
As to the image, take an arbitrary element of 
$$
\dfrac{N_1 N_2}{N_1} \times \dfrac{N_1 N_2}{N_2}
$$
and rewrite it 
$$
(n_{1} n_{2} N_{1}, n_{1}' n_{2}' N_{2})
=
(n_{2} N_{1}, n_{1}' N_{2})
=
(n_{1}' n_{2} N_{1}, n_{1}' n_{2} N_{2})
=
\varphi(n_{1}' n_{2}),
$$
where $n_{i}, n_{i}' \in N_{i}$.
A: You can assume $N_1\cap N_2=\{1\}$, by working in $G/(N_1\cap N_2)$ and using the fact that
$$
N_1N_2/N_1\cong(N_1N_2/(N_1\cap N_2))\big/(N_1/(N_1\cap N_2))
$$
and similary for the quotient modulo $N_2$. It is also not restrictive to assume $G=N_1N_2$.
In this case the isomorphism becomes the usual proof that, for a group $G$ with two normal subgroup $A$ and $B$ such that $AB=G$ and $A\cap B=\{1\}$, we have $G\cong A\times B$.
A: Here's a quick proof:
The second isomorphism theorem shows
$$ N_1N_2/(N_1)\cong N_2/(N_1\cap N_2) $$
and similarly for the other right-hand group.
So it would be enough to define a map $N_1\oplus N_2\rightarrow N_1N_2/(N_1\cap N_2)$ whose kernel is $N_1\cap N_2$.  That should be easy.
