An immediate consequence of this is that the subdirect product of two groups can be described as a fiber product and vice versa.
Notice that if is any subgroup of (the projections and need not be surjective), then the projections from onto and are surjective. Then one can apply Goursat's lemma to .
To motivate the proof, consider the slice in , for any arbitrary . By the surjectivity of the projection map to , this has a non trivial intersection with . Then essentially, this intersection represents exactly one particular coset of . Indeed, if we had distinct elements with and , then being a group, we get that , and hence, . But this a contradiction, as belong to distinct cosets of , and thus , and thus the element cannot belong to the kernel of the projection map from to . Thus the intersection of with every "horizontal" slice isomorphic to is exactly one particular coset of in .
By an identical argument, the intersection of with every "vertical" slice isomorphic to is exactly one particular coset of in .
All the cosets of are present in the group , and by the above argument, there is an exact 1:1 correspondence between them. The proof below further shows that the map is an isomorphism.
Before proceeding with the proof, and are shown to be normal in and , respectively. It is in this sense that and can be identified as normal in G and G', respectively.
Since is a homomorphism, its kernel N is normal in H. Moreover, given , there exists , since is surjective. Therefore, is normal in G, viz:
It follows that is normal in since
The proof that is normal in proceeds in a similar manner.
Given the identification of with , we can write and instead of and , . Similarly, we can write and , .
On to the proof. Consider the map defined by . The image of under this map is . Since is surjective, this relation is the graph of a well-defined function provided for every , essentially an application of the vertical line test.
Since (more properly, ), we have . Thus , whence , that is, .
Furthermore, for every we have . It follows that this function is a group homomorphism.
By symmetry, is the graph of a well-defined homomorphism . These two homomorphisms are clearly inverse to each other and thus are indeed isomorphisms.