# Kernel of group homomorphism $G\times G\to\mathbb{Z}$.

Consider a group epimorphism $$\varphi_1:G\to\mathbb{Z}$$. Is there any way to understand the kernel of the map $$\varphi_2:G\times G\to\mathbb{Z}$$ given by $$\varphi_2(g,h)=\varphi_1(g)-\varphi_1(h)=\varphi(gh^{-1})$$ in terms of $$\ker(\varphi_1)$$?

Obviously $$\ker(\varphi_1)\times \ker(\varphi_1)\subset\ker(\varphi_2)$$ and $$\Delta=\lbrace (g,g)\mid g\in G\rbrace\subset\ker(\varphi_2)$$ but there could be more elements since $$\varphi_1$$ need not be injective.

Is there any way to express $$\ker(\varphi_2)$$ in a closed form? (Maybe a semidirect product or something like that)

• Notice, that $(g,h)\in\ker(\varphi_2)\Leftrightarrow \varphi_1(g)=\varphi_1(h)$ (why?) Now which elements of G have the same image under $\varphi_1$? Jan 24 at 18:12
• @watertrainer $(g,h)\in\ker(\varphi_2)$ iff $gh^{-1}\in\ker(\varphi_1)$ iff $g\in h\ker(\varphi_1)$. Hence $(g,h)\in\ker(\varphi_2)$ iff they both lie in the same class of $G/\ker(\varphi_1)$. Right? Jan 24 at 18:22
• seems right to me, yes Jan 24 at 18:38
• @watertrainer and is there any way to wirte $\ker(\varphi_2)$ explicitely as a group?. I guess that maybe as a semidirect product of $\ker(\varphi_1)\times\ker(\varphi_1)$ but I,m not sure. Jan 24 at 19:00

Note that $$\ker(\varphi_2)$$ is a subdirect product of $$G\times G$$, since the projections onto each component are surjective. The subdirect products of $$A\times B$$ are classified by Goursat's Lemma: they are precisely the graphs of isomorphisms $$A/N\cong B/M$$ for normal subgroups $$N$$ of $$A$$ and $$M$$ of $$B$$.

Here, you get the graph of the identity map $$\frac{G}{\ker(\varphi_1)}\cong\frac{G}{\ker(\varphi_1)}$$. Thus, $$\ker(\varphi_2) = \bigcup_{g\in G} gN\times gN,$$ where $$N=\ker(\varphi_1)$$.

In particular:

Theorem. The elements of $$\ker(\varphi_2)$$ are the elements of the form $$(g,gn)$$ with $$n\in N=\ker(\varphi_1)$$.

Proof. All elements of the form $$(g,gn)$$ lie in $$\ker(\varphi_2)$$, since $$\varphi_1(g)-\varphi_1(gn) = \varphi_1(g)-\varphi_1(g) = 0$$.

Conversely, if $$(x,y)\in\ker(\varphi_2)$$, then $$\varphi_1(x)=\varphi_1(y)$$, so $$xN=yN$$. Thus, $$y=xn$$ for some $$n\in N$$, and $$(x,y)=(x,xn)$$. $$\Box$$

Note that this holds for any morphism into an abelian group $$f\colon G\to A$$, whether it is surjective or not.

• Thanks, that was what I needed. This proves that $\ker(\varphi_2)$ is finitely generated if $G$ is finitely generated, right? Jan 24 at 21:08
• @Marcos: Not sure about that. What if $N$ is not finitely generated? For example, take $G$ to be the free group of rank $2$ and $\varphi_1$ take $x$ to $1$ and $y$ to $0$. Then the kernel consists of all elements of the form $(g,gy^nc)$ with $g$ an arbitrary element, $n\in\mathbb{Z}$, and $c\in [G,G]$. But $[G,G]$ is not finitely generated, so I don't think you get $\ker(\varphi_2)$ finitely generated. Jan 24 at 21:58