# 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 and 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 $p_1$ and $N'$ the kernel of $p_2$. 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

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.

## Goursat varieties

As a consequence of Goursat's theorem, one can derive a very general version on the Jordan–HölderSchreier theorem in Goursat varieties.