Isomorphism between group mod inclusion and cartesian product mod normal subgroup

Question

Given $G_1, G_2$ Groups and $N$ a normal subgroup of $G_1\times G_2$, I have already proved that the projections $p_i: G_1\times G_2\to G_i$ and $i_1:g_1\mapsto (g_1,1), \; i_2:g_2\mapsto (1, g_2)$ are homomorphisms. Now, I need to prove that given $p_i(N)=G_i$ it holds that $$G_1/i_1^{-1}(N)\cong (G_1\times G_2)/N$$ My attempt: $G_1/i_1^{-1}(N)=\{gG_1 | g\in i_1^{-1}(N)\}$ and also $(G_1\times G_2)/N = \{g(G_1\times G_2)|g\in N\}$. I have tried proving that $i_1:G_1/i_1^{-1}(N)\to (G_1\times G_2)/N$ is bijective but the surjectivity does not work out. For this, I've also observed, that $i_1(i_1^{-1}(g))\in N \;\forall g\in N$.

Answer 1

To prove surjectivity observe you only need to prove if $(a_1,a_2) \in G_1\times G_2$ then $ \bar{(a_1,a_2)} = \bar{(g,1)}$ for some $g$. As $p_2(N) = G_2$, $(q,a_2) \in N$ for some $q$ now just take $g = a_1/q$