# 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 ${\displaystyle G}$, ${\displaystyle G'}$ be groups, and let ${\displaystyle H}$ be a subgroup of ${\displaystyle G\times G'}$ such that the two projections ${\displaystyle p_{1}:H\rightarrow G}$ and ${\displaystyle p_{2}:H\rightarrow G'}$ are surjective (i.e., ${\displaystyle H}$ is a subdirect product of ${\displaystyle G}$ and ${\displaystyle G'}$). Let ${\displaystyle N}$ be the kernel of ${\displaystyle p_{2}}$ and ${\displaystyle N'}$ the kernel of ${\displaystyle p_{1}}$. One can identify ${\displaystyle N}$ as a normal subgroup of ${\displaystyle G}$, and ${\displaystyle N'}$ as a normal subgroup of ${\displaystyle G'}$. Then the image of ${\displaystyle H}$ in ${\displaystyle G/N\times G'/N'}$ is the graph of an isomorphism ${\displaystyle G/N\approx G'/N'}$.

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

To motivate the proof, consider the slice ${\displaystyle S={g}\times G'}$ in ${\displaystyle G\times G'}$, for any arbitrary ${\displaystyle {g}\in G}$. By the surjectivity of the projection map to ${\displaystyle G}$, this has a non trivial intersection with ${\displaystyle H}$. Then essentially, this intersection represents exactly one particular coset of ${\displaystyle N'}$. Indeed, if we had distinct elements ${\displaystyle (g,a),(g,b)\in S\cap H}$ with ${\displaystyle a\in pN'\subset G'}$ and ${\displaystyle b\in qN'\subset G'}$ , then ${\displaystyle H}$ being a group, we get that ${\displaystyle (e,ab^{-1})\in H}$, and hence, ${\displaystyle (e,ab^{-1})\in N'}$. But this a contradiction, as ${\displaystyle a,b}$ belong to distinct cosets of ${\displaystyle N'}$, and thus ${\displaystyle ab^{-1}N'\neq N'}$, and thus the element ${\displaystyle (e,ab^{-1})\in N'}$ cannot belong to the kernel ${\displaystyle N'}$ of the projection map from ${\displaystyle H}$ to ${\displaystyle G}$. Thus the intersection of ${\displaystyle H}$ with every "horizontal" slice isomorphic to ${\displaystyle G'\in G\times G'}$ is exactly one particular coset of ${\displaystyle N'}$ in ${\displaystyle G'}$. By an identical argument, the intersection of ${\displaystyle H}$ with every "vertical" slice isomorphic to ${\displaystyle G\in G\times G'}$ is exactly one particular coset of ${\displaystyle N}$ in ${\displaystyle G}$.

All the cosets of ${\displaystyle G,G'}$ are present in the group ${\displaystyle H}$, 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, ${\displaystyle N}$ and ${\displaystyle N'}$ are shown to be normal in ${\displaystyle G\times \{e'\}}$ and ${\displaystyle \{e\}\times G'}$, respectively. It is in this sense that ${\displaystyle N}$ and ${\displaystyle N'}$ can be identified as normal in G and G', respectively.

Since ${\displaystyle p_{2}}$ is a homomorphism, its kernel N is normal in H. Moreover, given ${\displaystyle g\in G}$, there exists ${\displaystyle h=(g,g')\in H}$, since ${\displaystyle p_{1}}$ is surjective. Therefore, ${\displaystyle p_{1}(N)}$ is normal in G, viz:

${\displaystyle gp_{1}(N)=p_{1}(h)p_{1}(N)=p_{1}(hN)=p_{1}(Nh)=p_{1}(N)g}$.

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

${\displaystyle (g,e')N=(g,e')(p_{1}(N)\times \{e'\})=gp_{1}(N)\times \{e'\}=p_{1}(N)g\times \{e'\}=(p_{1}(N)\times \{e'\})(g,e')=N(g,e')}$.

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

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

On to the proof. Consider the map ${\displaystyle H\rightarrow G/N\times G'/N'}$ defined by ${\displaystyle (g,g')\mapsto (gN,g'N')}$. The image of ${\displaystyle H}$ under this map is ${\displaystyle \{(gN,g'N')|(g,g')\in H\}}$. Since ${\displaystyle H\rightarrow G/N}$ is surjective, this relation is the graph of a well-defined function ${\displaystyle G/N\rightarrow G'/N'}$ provided ${\displaystyle g_{1}N=g_{2}N\Rightarrow g_{1}'N'=g_{2}'N'}$ for every ${\displaystyle (g_{1},g_{1}'),(g_{2},g_{2}')\in H}$, essentially an application of the vertical line test.

Since ${\displaystyle g_{1}N=g_{2}N}$ (more properly, ${\displaystyle (g_{1},e')N=(g_{2},e')N}$), we have ${\displaystyle (g_{2}^{-1}g_{1},e')\in N\subset H}$. Thus ${\displaystyle (e,g_{2}'^{-1}g_{1}')=(g_{2},g_{2}')^{-1}(g_{1},g_{1}')(g_{2}^{-1}g_{1},e')^{-1}\in H}$, whence ${\displaystyle (e,g_{2}'^{-1}g_{1}')\in N'}$, that is, ${\displaystyle g_{1}'N'=g_{2}'N'}$.

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

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