# Goursat's lemma

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 theorem also implies the snake lemma.

## Groups

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

Let $G$ , $G'$  be groups, and let $H$  be a subgroup of $G\times G'$  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.

Notice that if $H$  is any subgroup of $G\times G'$  (the projections $p_{1}:H\rightarrow G$  and $p_{2}:H\rightarrow G'$  need not be surjective), then the projections from $H$  onto $p_{1}(H)$  and $p_{2}(H)$  are surjective. Then one can apply Goursat's lemma to $H\leq p_{1}(H)\times p_{2}(H)$ .

To motivate the proof, consider the slice $S={g}\times G'$  in $G\times G'$ , for any arbitrary ${g}\in G$ . 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 had distinct elements $(g,a),(g,b)\in S\cap H$  with $a\in pN'\subset G'$  and $b\in qN'\subset G'$  , then $H$  being a group, we get that $(e,ab^{-1})\in H$ , and hence, $(e,ab^{-1})\in N'$ . But this a contradiction, as $a,b$  belong to distinct cosets of $N'$ , and thus $ab^{-1}N'\neq N'$ , and thus the element $(e,ab^{-1})\in N'$  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 $G'\in G\times G'$  is exactly one particular coset of $N'$  in $G'$ . By an identical argument, the intersection of $H$  with every "vertical" slice isomorphic to $G\in G\times G'$  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 $G\times \{e'\}$  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 $\{(gN,g'N')|(g,g')\in H\}$ . Since $H\rightarrow G/N$  is surjective, this relation is the graph of a well-defined function $G/N\rightarrow G'/N'$  provided $g_{1}N=g_{2}N\Rightarrow g_{1}'N'=g_{2}'N'$  for every $(g_{1},g_{1}'),(g_{2},g_{2}')\in H$ , 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 $(g_{2}^{-1}g_{1},e')\in N\subset H$ . Thus $(e,g_{2}'^{-1}g_{1}')=(g_{2},g_{2}')^{-1}(g_{1},g_{1}')(g_{2}^{-1}g_{1},e')^{-1}\in H$ , whence $(e,g_{2}'^{-1}g_{1}')\in N'$ , that is, $g_{1}'N'=g_{2}'N'$ .

Furthermore, for every $(g_{1},g_{1}'),(g_{2},g_{2}')\in H$  we have $(g_{1}g_{2},g_{1}'g_{2}')\in H$ . It follows that this function is a group homomorphism.

By symmetry, $\{(g'N',gN)|(g,g')\in H\}$  is the graph of a well-defined homomorphism $G'/N'\rightarrow G/N$ . 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–HölderSchreier theorem in Goursat varieties.