Stallings' Group is Simply Connected at Infinity

Michael Mihalik

Published 2025-06-23Version 1

Let $F_2$ be the free group on two generators and let $B_n$ ($n \geq 2$) denote the kernel of the homomorphism $$F_2 \times \cdots (n) \cdots \times F_2 \rightarrow {\mathbb Z}$$ sending all generators to the generator $1$ of $\mathbb Z$. The groups $B_k$ are called the {\it Bieri-Stallings} groups and $B_k$ is type $\mathcal F_{k-1}$ but not $\mathcal F_k$. For $n\geq 3$ there are short exact sequences of the form $$1 \rightarrow B_{n-1} \rightarrow B_n \rightarrow F_2 \rightarrow 1.$$ This exact sequence can be used to show that $B_n$ is $(n-3)$-connected at infinity for $n\geq 3$. Stallings' proved that $B_2$ is finitely generated but not finitely presented. We conjecture that for $n\geq 2$, $B_n$ is $(n-2)$-connected at infinity. For $n=2$, this means that $B_2$ is 1-ended and for $n=3$ that $B_3$ (typically called Stallings' group) is simply connected at infinity. We verify the conjecture for $n=2$ and $n=3$. Our main result is the case $n=3$: Stalling's group is simply connected at $\infty$.