so here is what I have:

$G$ is a group. For $i \in \{1,2\}: p_i:G \twoheadrightarrow G_i$ is a surjective morphism of groups. $H_i = ker(p_i), H_1 \cap H_2 = \{1\}$

In the other parts of the question I found out that:

$p: G \rightarrow G_1 \times G_2$ given by $p(g) = (p_1(g),p_2(g))$ is injective.

$K_1 = p_1(H_2) = p_1(H)\lhd G_1$ and $K_2 = p_2(H_1)= p_2(H)\lhd G_2$ and $H= H_1 \cdot H_2 \lhd G$ and $p(H) = K_1 \times K_2$. I also know that $p^{-1}_1(K_1)= H$ and $p_2^{-1}(K_2)=H$ and there are isomorphisms of groups $\bar{p_i}: G/H \rightarrow G_i/K_i$ sending $gH \mapsto p_i(g)\cdot K_i$

From this I am supposed to find the isomorphism induced by $p$ that maps $G \rightarrow L=\{(x_1,x_2) \in G_1 \times G_2 | f_1(x_1)=f_2(x_2)\}$ where $f_i: G_i \rightarrow G/H, f_i=\bar{p_i}^{-1} \circ \pi: G_i \overset{\pi}\twoheadrightarrow G_i/K_i \overset{\bar{p_i}^{-1}} \rightarrow G/H$

I figured that there would be a function $\bar{p}: G/H \rightarrow (G_1/K_1, G_2/K_2)$ mapping $gH \mapsto (p_1(g)\cdot K_1, p_2(g)\cdot K_2)$ and that $f_1(x_1) = f_2(x_2) \iff p_1(g) = p_2(g)$ for some g.

I'm stuck at this point and just can't figure out how to proceed. Any help would really be greatly appreciated!


The solution from class in the end was this:

$ p: G\rightarrow L=\{(x_1,x_2) \in G_1 \times G_2 | f_1(x_1)=f_2(x_2)\} \subset G_1\times G_2$

because we know that $p$ is an injective morphism of groups and for

$g\in G: p(g)=(p_1(g),p_2(g))$ and $f_1(p_1(g)) = \bar{p_1}^{-1} \circ \pi (p_1(g)) = \pi(g) = f_2(p_2(g)) \Rightarrow p(g)\in L.$

Also for $(x_1,x_2)\in L, x_i := p_i(g_i), g_i\in G:$

$\pi(g_1) = \bar{p_1}^{-1}\circ\pi\circ p_1(g_1) = f_1\circ p_1(g_1) = f_1(x_1) = f_2(x_2) = \pi(g_2) $

$\Rightarrow g_1^{-1}g_2 \in H = ker(\pi) = H_1\cdot H_2 \Rightarrow g_1^{-1}g_2=h_1h_2, h_i\in H_i, g:=g_1h_1 =g_2h_2^{-1},$

$\Rightarrow p(g) = (p_1(g),p_2(g)) = (p_1(g_1h_1), p_2(g_2h_2^{-1})) = (p_1(g_1), p_2(g_2)) = (x_1,x_2)$


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