Isodynamic point

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

It 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.

Proof of Goursat's lemma

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.

References

• Kenneth A. Ribet (Autumn 1976), "Galois Action on Division Points of Abelian Varieties with Real Multiplications", American Journal of Mathematics, Vol. 98, No. 3, 751–804.