Prove: $G/(N_1 \cap N_2)$ is isomorphic with a subgroup of $(G/N_1) \times (G/N_2)$ I have to solve this exercise for my math study, but don't know how to do it. It's keeping me busy for 2 days now.
Let $G$ be a group and $N_{1}, N_{2}$ normal subgroups of $G$. Let $f: G \rightarrow (G/N_{1}) \times (G/N_{2})$ with $f(a) = (aN_{1}, aN_{2})$be a homomorphism with $Ker(f) = N_{1}\cap N_{2}$.
Prove that $G/N_{1}\cap N_{2}$ is isomorphic to a subgroup of $(G/N_{1}) \times (G/N_{1})$.
So far, I have done this:
$f$ is an homomorphism and $Ker(f) = N_{1}\cap N_{2}$, so $N_{1}\cap N_{2}$ is a normal subgroup of $G$. I also think that I have to use the fundamental homomorphism theorem, but I don't know how.
Could you please tell me how to prove this? I have really tried to solve this for a very long time, but I don't see what I have to do.
Thanks in advance!
 A: There are a lot of unnecessary distractions in this problem. In general, if $f:G\to H$ is a homomorphism with kernel $K$, then the image of $f$, which is a subgroup of $H$, is isomorphic to $G/K$. This follows from the first isomorphism theorem.
A: When you say "Let $G$ be a group and $N_{1}, N_{2}$ normal subgroups of $G$. Let $f: G \rightarrow (G/N_{1}) \times (G/N_{1})$ be a homomorphism with $Ker(f) = N_{1}\cap N_{2}$", then you still have to define your homomorphism. This should be $f(g)=(gN_1, gN_2).$Then you have to prove that (a) this $f$ is a homomorphism and (b) that $ker(f)=N_1 \cap N_2$. Can you proceed from here?
A: If you prove that the map $f$ is surjective, then you have the claim. Indeed, by the first isomorphism theorem you have that $Im(f) \cong (G/N_1) \times (G/N_2)$.
A: $\textbf{Def:}$ Let $A,B$ be groups. Then, $A \times B=\{(a,b)|a \in A, b \in B\}$. $A \times B$ is a group by defining $(a_1,b_1) \cdot (a_2,b_2) \,\, \text{for} \,\, a_1,a_2 \in A \,\, \text{and} \,\, b_1,b_2 \in B$ called the $\textbf{Direct Product}$ of $A$ and $B$. (You can check that this is a group by checking closure, identity, associativity, and inverse).
$\textbf{Proof:}$ Define $\varphi:G \rightarrow G/N_1 \times G/N_2$ by $\varphi(G)=(N_1g,N_2g) \,\, \forall g \in G$. 
So, $\text{ker}(\varphi)=\{g \in G | \varphi(G)=1_{G/N_1 \times G/N_2}\}=\{g \in G | \varphi(G)=(N_1,N_2)\}=\{g \in G | (N_1g,N_2g)=(N_1,N_2)\}$. 
We know that $N_1g=N_1$ if and only if $g \in N$.

Thus, $\text{ker}(\varphi)=\{g \in G | g \in N_1 and g \in N_2\} = \{ g \in G | g \in N_1 \cap N_2\} = N_1 \cap N_2$.

Then, by the First Isomorphism Theorem,
$G/N_1 \cap N_2 \cong \varphi(G)$.
$\,\, \blacksquare$

(To be more rigorous, you can show that $\varphi$ is a homomorphism).
