Surjectivity of $G \rightarrow G/N_1 \times G/N_2$. Given a group $G$ with $N_1$, $N_2$ normal in $G$ such that $(G:N_1)$ and $(G:N_2)$ are coprime (in particular, $N_1,N_2$ have finite index in $G$). How do you prove that the natural map $G \rightarrow G/N_1 \times G/N_2$ is surjective?
 A: Look at the value of $(G: N_1 \cap N_2)$. This value is divisible by both $(G: N_1)$ and $(G: N_2)$. Since they are coprime, $(G: N_1 \cap N_2) \ge (G: N_1) \cdot (G: N_2)$.
Now, the natural map $\varphi \colon G/(N_1 \cap N_2) \to G/N_1 \times G/N_2$ is clearly injective, and the group on the left has at least as many elements as the group on the right. It follows that $\varphi$ is surjective, and so is the homomorphism in the original question.
A: I was trying to give you another different approach, but what I could prepare, after hours, contains little differences with @Dan's. There is an stared exercise in  J.J.Rose's book in Group Theory which says:

Let $H,K$ are subgroups of group (not necessarily finite) $G$ such that they are of finite co-prime index in $G$. Therefore $$[G:H\cap K]=[G:H][G:K]$$ and $G=HK$.

We have the homomorphism $$\phi:G \rightarrow G/H \times G/K\\\phi(g)=(gH,gK)$$ and wants to prove it is onto. Take an element of RHS, say $(g_1H,g_2K)$. By $G=HK$ and then every element of the group can be written as the product of an element of $H$ by an element of $K$. In fact $$g_1=h_1k_1,~~g_2=h_2k_2,~~ h_i\in H,~ k_i\in K$$ It is enough to evaluate $\phi(k_1h_2)$.
