Goursat's lemma

Groups

Goursat's lemma for groups can be stated as follows.

Let G , G ′ be groups, and let H be a subgroup of G × G ′ such that the two projections p 1 : H → G and p 2 : H → G ′ are surjective (i.e., H is a subdirect product of G and G ′ ). Let N be the kernel of p 2 and N ′ the kernel of p 1 . One can identify N as a normal subgroup of G , and N ′ as a normal subgroup of G ′ . Then the image of H in G / N × G ′ / N ′ is the graph of an isomorphism G / N ≈ G ′ / N ′ .

An immediate consequence of this is that the subdirect product of two groups can be described as a fiber product and vice versa.

To motivate the proof, consider the slice S = g × G ′ in G × G ′ , for any arbitrary g ∈ G . By the surjectivity of the projection map to G , this has a non trivial intersection with H . Then essentially, this intersection represents exactly one particular coset of G ′ . Indeed, if we had distinct elements ( g , a ) , ( g , b ) ∈ S with a ∈ p N ′ ⊂ G ′ and a ∈ q N ′ ⊂ G ′ , then H being a group, we get that ( e , a b − 1 ) ∈ H , and hence, ( e , a b − 1 ) ∈ N ′ . But this a contradiction, as a , b belong to distinct cosets of N ′ , and thus a b − 1 N ′ ≠ N ′ , and thus the element ( e , a b − 1 ) ∈ N ′ cannot belong to the kernel N ′ of the projection map from H to G . Thus the intersection of H with every "horizontal" slice isomorphic to G ′ ∈ G × G ′ is exactly one particular coset of N ′ in G ′ . By an identical argument, the intersection of H with every "vertical" slice isomorphic to G ∈ G × G ′ is exactly one particular coset of N in G .

All the cosets of G , G ′ are present in the group H , 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.

Proof

Before proceeding with the proof, N and N ′ are shown to be normal in G × { e ′ } and { e } × G ′ , respectively. It is in this sense that N and N ′ can be identified as normal in G and G', respectively.

Since p 2 is a homomorphism, its kernel N is normal in H. Moreover, given g ∈ G , there exists h = ( g , g ′ ) ∈ H , since p 1 is surjective. Therefore, p 1 ( N ) is normal in G, viz:

g p 1 ( N ) = p 1 ( h ) p 1 ( N ) = p 1 ( h N ) = p 1 ( N h ) = p 1 ( N ) g .

It follows that N is normal in G × { e ′ } since

( g , e ′ ) N = ( g , e ′ ) ( p 1 ( N ) × { e ′ } ) = g p 1 ( N ) × { e ′ } = p 1 ( N ) g × { e ′ } = ( p 1 ( N ) × { e ′ } ) ( g , e ′ ) = N ( g , e ′ ) .

The proof that N ′ is normal in { e } × G ′ proceeds in a similar manner.

Given the identification of G with G × { e ′ } , we can write G / N and g N instead of ( G × { e ′ } ) / N and ( g , e ′ ) N , g ∈ G . Similarly, we can write G ′ / N ′ and g ′ N ′ , g ′ ∈ G ′ .

On to the proof. Consider the map H → G / N × G ′ / N ′ defined by ( g , g ′ ) ↦ ( g N , g ′ N ′ ) . The image of H under this map is { ( g N , g ′ N ′ ) | ( g , g ′ ) ∈ H } . Since H → G / N is surjective, this relation is the graph of a well-defined function G / N → G ′ / N ′ provided g 1 N = g 2 N ⇒ g 1 ′ N ′ = g 2 ′ N ′ for every ( g 1 , g 1 ′ ) , ( g 2 , g 2 ′ ) ∈ H , essentially an application of the vertical line test.

Since g 1 N = g 2 N (more properly, ( g 1 , e ′ ) N = ( g 2 , e ′ ) N ), we have ( g 2 − 1 g 1 , e ′ ) ∈ N ⊂ H . Thus ( e , g 2 ′ − 1 g 1 ′ ) = ( g 2 , g 2 ′ ) − 1 ( g 1 , g 1 ′ ) ( g 2 − 1 g 1 , e ′ ) − 1 ∈ H , whence ( e , g 2 ′ − 1 g 1 ′ ) ∈ N ′ , that is, g 1 ′ N ′ = g 2 ′ N ′ .

Furthermore, for every ( g 1 , g 1 ′ ) , ( g 2 , g 2 ′ ) ∈ H we have ( g 1 g 2 , g 1 ′ g 2 ′ ) ∈ H . It follows that this function is a group homomorphism.

By symmetry, { ( g ′ N ′ , g N ) | ( g , g ′ ) ∈ H } is the graph of a well-defined homomorphism G ′ / N ′ → G / N . These two homomorphisms are clearly inverse to each other and thus are indeed isomorphisms.

Goursat varieties

As a consequence of Goursat's theorem, one can derive a very general version on the Jordan–Hölder–Schreier theorem in Goursat varieties.