Prove $N_1N_2/(N_1\cap N_2)\cong (N_1N_2/N_1) \times (N_1N_2/N_2)$, where $N_1, N_2$ are normal subgroups of $G$ Prove the following statement:
$N_1N_2/(N_1\cap N_2)\cong (N_1N_2/N_1) \times (N_1N_2/N_2)$
where $N_1, N_2$ are normal subgroups of $G$
How do I approach this question? I tried using the first isomorphism theorem, but got stuck. Maybe I should use a different approach.
 A: There is a canonical quotient map $N_1N_2\to N_1N_2/N_i$ for each $i$. If there is any good in this world, then the resulting product map
$$
f:N_1N_2\to(N_1N_2/N_1) \times (N_1N_2/N_2)
$$
is surjective and has $N_1\cap N_2$ as kernel, which by the first isomorphism theorem gives you the isomorphism you're looking for.
The kernel is easy to find. For any $n\in N_1N_2$, the first component of $f(n)$ is the identity iff $n\in N_1$, and the second component is the identity iff $n\in N_2$. So $f(n)$ is the identity iff $n\in N_1\cap N_2$.
Now for surjectivity (and way too many symbols all using the 14th letter of the alphabet, sorry about that). Take an arbitrary $(n_1N_1, n_2N_2)\in (N_1N_2/N_1) \times (N_1N_2/N_2)$. We note that
$$
f(n_1) = (n_1N_1, n_1N_2)\\
f(n_1)(eN_1, n_1^{-1}n_2N_2) = (n_1N_1, n_2N_2)
$$
So we have reduced the question to this: Given any $n'\in N_1N_2$, can we find an $n_3\in N_1N_2$ such that $f(n_3) = (eN_1, n'N_2)$? For if that is the case, then with $n' = n_1^{-1}n_2$, we will have $f(n_1n_3) = (n_1N_1, n_2N_2)$.
Note that $n'\in N_1N_2$, which means we have $n' = n'_1n_2'$ with $n_1'\in N_1, n_2'\in N_2$. And $n_1'n_2'N_2 = n_1'N_2$. So we may set $n_3 = n_1'$, and we are done.
