# Goursat's lemma

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

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.

### 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.