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)

  • $\begingroup$ 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 $\varphi_1$? $\endgroup$ Jan 24 at 18:12
  • 2
    $\begingroup$ @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? $\endgroup$
    – Marcos
    Jan 24 at 18:22
  • $\begingroup$ seems right to me, yes $\endgroup$ Jan 24 at 18:38
  • $\begingroup$ @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. $\endgroup$
    – Marcos
    Jan 24 at 19:00

1 Answer 1


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.

  • $\begingroup$ Thanks, that was what I needed. This proves that $\ker(\varphi_2)$ is finitely generated if $G$ is finitely generated, right? $\endgroup$
    – Marcos
    Jan 24 at 21:08
  • $\begingroup$ @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. $\endgroup$ Jan 24 at 21:58

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