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!


1 Answer 1


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) $

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

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .