# Goursat's lemma

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

Goursat's lemma is an algebraic theorem about subgroups of the direct product of two groups.

It 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 of Goursat's lemma

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.

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 ${\displaystyle \{(gN,g'N')|(g,g')\in H\}}$. This relation is the graph of a well-defined function ${\displaystyle G/N\rightarrow G'/N'}$ provided ${\displaystyle gN=N\Rightarrow g'N'=N'}$, essentially an application of the vertical line test.

Since ${\displaystyle gN=N}$ (more properly, ${\displaystyle (g,e')N=N}$), we have ${\displaystyle (g,e')\in N\subset H}$. Thus ${\displaystyle (e,g')=(g,g')(g^{-1},e')\in H}$, whence ${\displaystyle (e,g')\in N'}$, that is, ${\displaystyle g'N'=N'}$. Note that by symmetry, it is immediately clear that ${\displaystyle g'N'=N'\Rightarrow gN=N}$, 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.