Proving that a countable group is not finitely generated

Question

I would like to learn about techniques for proving that a countable group is not finitely generated. I am also interested in learning about examples. Finally, I am particularly, but not exclusively, interested in topological methods. Thank you very much!

Answer 1

The easiest trick, if it happens to work, is to construct a surjection onto another group that you already know is not finitely generated. A canonical quotient to try is the abelianization.

Answer 2

Following Giles Gardam's answer, the second easiest trick is to say that a countable group $G$ is fintely generated if and only every sequence of subgroups $H_1 \subset H_2 \subset \cdots$ satisfying $G= \bigcup_{i \geq 1} H_i$ is eventually constant (to $G$). For instance, $\mathrm{FAlt}(\mathbb{N})$ - namely, the group of finitely supported even permutations of $\mathbb{N}$ - is simple, so we cannot use the trick of surjecting it onto an infinitely generated group we already know, but we have the increasing sequence in it given by the $\mathrm{Alt}([0,n])$.

Answer 3

The main topological method I know is the following special case of Brown's criterion (see Theorem 2.2 of http://pi.math.cornell.edu/~kbrown/scan/1987.0044.0045.pdf, using $n=1$): Suppose a group $G$ acts on a connected graph $X$ such that every vertex stabilizer is finitely generated (for example maybe they're finite, or even trivial). Assume there is a filtration of $X$ into $G$-invariant, cocompact (meaning finitely many orbits of vertices and edges) subgraphs $X_1 \subseteq X_2 \subseteq X_3 \subseteq \cdots$ such that for all $i<j$ there exist vertices $x$ and $y$ in $X_i$ that lie in different connected components of $X_j$. Then $G$ is not finitely generated. (You can also have the subgraphs indexed not by $\mathbb{N}$ but by any directed set $I$, still using this "for all $i<j$" thing.)

As for an actual example that uses this, I guess the first thing that comes to mind is, one can use this to recover Nagao's result that $SL_2(\mathbb{Z}[t])$ is not finitely generated, see Bux--Mohammadi--Wortman: https://ems.press/content/serial-article-files/3521 (specialized to $n=2$), especially Subsection 4.4, "Applying Brown's criterion" (Although this is sort of overkill in the $n=2$ case.)

Answer 4

One possible way of proving that a countable group $G$ is not finitely generated is finding an infinite set $S$ and a mapping $\varphi$ from $G$ to the power set of $S$ such that the following hold:

• $\forall g \in G \ |\varphi(g)| < \infty$,

• $\forall g, h \in G \ \varphi(gh) \subseteq \varphi(g) \cup \varphi(h)$, and

• $\cup_{g \in G} \varphi(g) = S$.

This method can be used e.g. to prove that the group ${\rm CT}(\mathbb{Z})$ generated by all class transpositions is not finitely generated. --

Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

In this case, $S$ is the set of prime numbers — and given $g \in {\rm CT}(\mathbb{Z})$, the set $\varphi(g)$ is the prime set of $g$. That is, it consists of the prime numbers which divide at least one of the coefficients $a_{r(m)}$ or $c_{r(m)}$ of the mappings $g|_{r(m)}: n \mapsto (a_{r(m)} n + b_{r(m)})/c_{r(m)}$, where the $r(m)$ are chosen such that the restrictions indeed have that form.