Goursat's lemma

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Goursat's lemma, named after the French mathematician Édouard Goursat, is an algebraic theorem about subgroups of the direct product of two groups.

It can be stated more generally in a Goursat variety (and consequently it also holds in any Maltsev variety), from which one recovers a more general version of Zassenhaus' butterfly lemma. In this form, Goursat's lemma also implies the snake lemma Template:Citation needed.

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 p1:H→G and p2:H→G′ are surjective (i.e., H is a subdirect product of G and G′). Let N be the kernel of p2 and N′ the kernel of p1. 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′. One then obtains a bijection between:
  1. Subgroups of G×G′ which project onto both factors,
  2. Triples (N,N′,f) with N normal in G, N′ normal in G′ and f isomorphism of G/N onto 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.

Notice that if H is any subgroup of G×G′ (the projections p1:H→G and p2:H→G′ need not be surjective), then the projections from H onto p1(H) and p2(H) are surjective. Then one can apply Goursat's lemma to H≤p1(H)×p2(H).

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 N′. Indeed, if we have elements (g,a),(g,b)∈S∩H with a∈pN′⊂G′ and b∈qN′⊂G′, then H being a group, we get that (e,ab−1)∈H, and hence, (e,ab−1)∈N′. It follows that (g,a) and (g,b) lie in the same coset of N′. 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 N,N′ 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 p2 is a homomorphism, its kernel N is normal in H. Moreover, given g∈G, there exists h=(g,g′)∈H, since p1 is surjective. Therefore, p1(N) is normal in G, viz:

gp1(N)=p1(h)p1(N)=p1(hN)=p1(Nh)=p1(N)g.

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

(g,e′)N=(g,e′)(p1(N)×{e′})=gp1(N)×{e′}=p1(N)g×{e′}=(p1(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 gN 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′)↦(gN,g′N′). The image of H under this map is {(gN,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 g1N=g2N⟹g1′N′=g2′N′ for every (g1,g1′),(g2,g2′)∈H, essentially an application of the vertical line test.

Since g1N=g2N (more properly, (g1,e′)N=(g2,e′)N), we have (g2−1g1,e′)∈N⊂H. Thus (e,g2'−1g1′)=(g2,g2′)−1(g1,g1′)(g2−1g1,e′)−1∈H, whence (e,g2'−1g1′)∈N′, that is, g1′N′=g2′N′.

Furthermore, for every (g1,g1′),(g2,g2′)∈H we have (g1g2,g1′g2′)∈H. It follows that this function is a group homomorphism.

By symmetry, {(g′N′,gN)∣(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

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

References