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## The easiest special case of the main theorem about BNS invariants

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.