It can be stated as follows.
- Let , be groups, and let be a subgroup of such that the two projections and are surjective (i.e., is a subdirect product of and ). Let be the kernel of and the kernel of . One can identify as a normal subgroup of , and as a normal subgroup of . Then the image of in is the graph of an isomorphism .
An immediate consequence of this is that the subdirect product of two groups can be described as a fiber product and vice versa.
Proof of Goursat's lemma
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 , .
Since (more properly, ), we have . Thus , whence , that is, . Note that by symmetry, it is immediately clear that , i.e., this function also passes the horizontal line test, and is therefore one-to-one. The fact that this function is a surjective group homomorphism follows directly.