Let \(G\) be a finitely generated group. Let \(S(G)\) be the set of equivalence classes of characters \(G \rightarrow \mathbb{R}\) modulo rescaling by positive real numbers. Recall that the **BNS invariant** of \(G\) is the set \(\Sigma(G)\) of all \([\rho] \in S(G)\) such that the full subgraph of the Cayley graph of \(G\) spanned by \(g \in G\) such that \(\rho(g) \geq 0\) is connected. This does not depend on the generating set (see this MathOverflow question). The main theorem about the BNS invariants is the following:

**Theorem**: Let \(G\) be a finitely generated group and let \(H\) be a subgroup of \(G\) such that \([G,G] \subset H\). Then \(H\) is finitely generated if and only if \([\rho] \in \Sigma(G)\) for all characters \(\rho\colon G \rightarrow \mathbb{R}\) that vanish on \(H\).

The purpose of this post is to give a short proof of the following very special case of this:

**Theorem:** Let \(G\) be a finitely generated group and let \(\rho\colon G \rightarrow \mathbb{Z}\) be a surjective homomorphism. Set \(H = \text{ker}(\rho)\), and assume that \([\rho],[-\rho] \in \Sigma(G)\). Then \(H\) is finitely generated.

Pick \(t \in G\) such that \(\rho(t) = 1\). Since \(G\) is finitely generated, we can pick a finite subset \(S \subset H\) such that \(S \cup \{t\}\) generates \(G\). Of course, \(S\) need not generate \(H\). However, it has to at least normally generate \(H\), so \(H\) is generated by the set

$$\bigcup_{k=-\infty}^{\infty} t^k S t^{-k}.$$

A moment’s though will show that the fact that \([\rho] \in \Sigma(G)\) is equivalent to the assertion that in fact \(H\) is generated by

$$\bigcup_{k=0}^{\infty} t^k S t^{-k}.$$

Similarly, the fact that \([-\rho] \in \Sigma(G)\) is equivalent to the assertion that \(H\) is generated by

$$\bigcup_{k=-\infty}^0 t^k S t^{-k}.$$

We want to parley these two fact into showing that \(H\) is finitely generated.

**Remark**: If \(H\) were finitely generated, then it would be generated by a finite set of conjugates of elements of \(S\) by powers of \(t\). Since \(H\) is a normal subgroup, we can conjugate any generating set for it by any power of \(t\) that we like and it remains a generating set. From this, we see that in fact there would exist some \(N \geq 0\) such that \(H\) is generated by both

$$\bigcup_{k=0}^N t^k S t^{-k} \quad \quad \text{and} \quad \quad \bigcup_{k=-N}^0 t^k S t^{-k}.$$

This shows the necessity for finite generation of \([\rho],[-\rho] \in \Sigma(G)\).

We now return to the proof. Since \(H\) is generated by

$$\bigcup_{k=-\infty}^0 t^k S t^{-k},$$

for all \(s \in S\) we can write \(t s t^{-1}\) as a product of conjugates of elements \(S\) by negative powers of \(t\). We deduce that there exists some large \(N \geq 0\) such that the subgroup generated by the set

$$S’ := \bigcup_{k=-N}^0 t^k S t^{-k}$$

contains \(t s t^{-1}\) for all \(s \in S\). We claim that \(S’\) generates \(H\), i.e. that the subgroup \(H’\) of \(H\) generated by \(S’\) satisfies \(H’ = H\). Before we do this, we verify the following claim:

**CLAIM**: \(t H’ t^{-1} \subset H’\)

The group \(t H’ t^{-1}\) is generated by the set

$$\bigcup_{k=-N+1}^1 t^k S t^{-k}.$$

The only elements of this that are not already elements of \(S’\) are the elements \(t s t^{-1}\) for \(s \in S\); however, we chose \(S’\) precisely so that these are all products of elements of \(S’\). We conclude that \(t H’ t^{-1} \subset H’\), as claimed.

We now prove that \(H’ = H\). Since \(H’ \subset H\), it is enough to prove the reverse inclusion. By iterating the above claim, we see that for all \(k \geq 0\) we have \(t^{k} H’ t^{-k} \subset H’\). Since \(S \subset H’\), this implies that \(t^k S t^{-k} \subset H’\). Our assumption that \([\rho] \in \Sigma(G)\) (which we still have not used!) is equivalent to the fact that \(H\) is generated by the set

$$\bigcup_{k=0}^{\infty} t^k S t^{-k}.$$

Since all these generators lie in \(H’\), we conclude that \(H \subset H’\), as desired.