$$\kappa(G,H):=\cap_{\phi \in {\rm Hom}(G,H)} \ker\phi$$ where ${\rm Hom}(G,H)$ is the set of all homomorphisms from $G$ to $H$.

I am trying to prove that if $G$ is a finitely generated group, and $H$ a finite group, then $[G:\kappa(G,H)]<\infty$.

Since I had no idea on how to start proving this generally, I tried proving a simplified version of this, where both $G$ and $H$ are abelian. In this case, I can use the Fundamental Theorem of Finitely Generated Abelian Groups and assume that $$G = \mathbb{Z}/(g_1)\oplus\mathbb{Z}/(g_2)\oplus\dots\mathbb{Z}/(g_k)\oplus\mathbb{Z}^r$$

$$H = \mathbb{Z}/(h_1)\oplus\mathbb{Z}/(h_2)\oplus\dots\oplus\mathbb{Z}/(h_m)$$

I then observed that for all homomorphisms $\phi:G\to H$ we have that $\ker\phi$ has the form: $$N_1 \oplus N_2 \oplus\dots\oplus N_k \oplus m_1\mathbb{Z}\oplus\dots\oplus m_r\mathbb{Z}$$ for some finite normal subgroups $N_i \leq \mathbb{Z}/(g_i)$ and normal subgroups $m_1\mathbb{Z}\oplus\dots\oplus m_r\mathbb{Z} \leq \mathbb{Z}^r$ (we can see this by defining each possible homomorphism using the standard generators (e.g., $(0,\dots,0,1,0\dots0)$). Which means that when we take the intersection of all these kernels, we still get something of that form (e.g., instead of $m_1\mathbb{Z}$ we will get the ${\rm lcm}$ of all $m_1$'s taken across all $\phi$'s), so that $[G:\kappa(G,H)]<\infty$).

I am, however, none the wiser about how to generalize this to the non-abelian case.

I do have at my disposal two potentially related facts (although I do not see how they might be helpful... so I may be completely off here):

  1. For any two groups $G, H$, we have that $H$ is isomorphic to a quotient of $G$ iff $H$ is isomorphic to a quotient of $G/\kappa(G,H)$.
  2. For any groups $G, G', H$ we have: $\kappa(G\times G', H) = \kappa(G,H)\times\kappa(G', H)$

Any advice?

  • 2
    $\begingroup$ Are you able to prove that if $A, B$ are finite index subgroups of your group $G$, then $A\cap B$ also has finite index? And can you then see why this is relevant? $\endgroup$
    – user1729
    Jan 20, 2023 at 15:09
  • 2
    $\begingroup$ Here's another fact: for any two groups $G,H$ and any homomorphism $\phi : G \to H$, the index of $\text{ker}(\phi)$ equals the order of $\text{image}(\phi)$. $\endgroup$
    – Lee Mosher
    Jan 20, 2023 at 15:39

1 Answer 1


(Thanks to user1729 and Lee Mosher)

It is sufficient to prove that $[G:\ker\phi]<\infty$ for all homomorphisms $\phi: G \to H$, since:

  1. If $A,B$ are finite index subgroups of $G$ then $A \cap B$ is also of finite index. This can be generalized to a finite number of intersections.
  2. Since $H$ is finite, we have a finite number of homomorphisms $n$, which means no. 1 applies to $\kappa(G,H) = \cap_i^n \ker\phi_i$.

Indeed, let $\phi: G \to H$ be a homomorphism. Then by the first isomorphism theorem, we have that $[G:\ker\phi] = |{\rm Im}\, g(\phi)|<|H|<\infty$.

(as a side note - one way in which I could have realized how to generalize the abelian case was to think of which part of the observation:

"Which means that when we take the intersection of all these kernels, we still get something of that form"

was essential and generalizable, and that is that the intersection retains the essential property of being of finite index [and stripping the non-essential parts, e.g. of being in $\mathbb{Z}$]).


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