Goursat's lemma

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Not to be confused with Goursat's integral lemma from Complex analysis

Goursat's lemma is an algebraic theorem about subgroups of the direct product of two groups.

It can be stated as follows.

Let G, G' be groups, and let H be a subgroup of G\times G' such that the two projections p_1: H\rightarrow G and p_2: H\rightarrow 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\times G'/N' is the graph of an isomorphism G/N\approx 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.

Proof of Goursat's lemma[edit]

Before proceeding with the proof, N and N' are shown to be normal in G \times \{e'\} and \{e\} \times 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 \in G, there exists h=(g,g') \in H, since p_1 is surjective. Therefore, p_1(N) is normal in G, viz:

gp_1(N)=p_1(h)p_1(N)=p_1(hN)=p_1(Nh)=p_1(N)g.

It follows that N is normal in G \times \{e'\} since

(g,e')N = (g,e')(p_1(N) \times \{e'\}) = gp_1(N) \times \{e'\} = p_1(N)g \times \{e'\} = (p_1(N) \times \{e'\})(g,e')=N(g,e').

The proof that N' is normal in \{e\} \times G' proceeds in a similar manner.

Given the identification of G with G \times \{e'\}, we can write G/N and gN instead of (G \times \{e'\})/N and (g,e')N, g \in G. Similarly, we can write G'/N' and g'N', g' \in G'.

On to the proof. Consider the map H \rightarrow G/N \times G'/N' defined by (g,g') \mapsto (gN, g'N'). The image of H under this map is \{(gN,g'N') | (g,g') \in H \}. This relation is the graph of a well-defined function G/N \rightarrow G'/N' provided gN=N \Rightarrow g'N'=N', essentially an application of the vertical line test.

Since gN=N (more properly, (g,e')N=N), we have (g,e') \in N \subset H. Thus (e,g') = (g,g')(g^{-1},e') \in H, whence (e,g') \in N', that is, g'N'=N'. Note that by symmetry, it is immediately clear that g'N'=N' \Rightarrow gN=N, 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.

References[edit]