Goursat's lemma

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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.

[edit] Proof of Goursat's Lemma

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.