8
$\begingroup$

I'm looking for an example of a finitely presented and finitely generated amenable group, that has a subgroup which is not finitely generated.

The question is easy for finitely generated amenable group and an example is the lamp-lighter group $C_2\wr \mathbb{Z}$.

An Abelian and finitely generated group has no such subgroups. There exists a bigger class of groups with this property?

$\endgroup$

3 Answers 3

13
$\begingroup$

I don't know much about amenable groups I am afraid, but according to the Wikipedia article, all solvable groups are amenable. So we can take the Baumslag-Solitar group

$B(1,n) = \langle x,y \mid y^{-1}xy = x^n \rangle.$

If we let $N$ be the normal closure of the subgroup generated by $x$, then $N$ is abelian with $G/N$ cyclic, but $N$ is not finitely generated when $n > 1$. Note also that $B(1,n)$ is isomorphic to the subgroup of ${\rm GL}(2, \mathbb{Q})$ generated by

$x = \left(\begin{array}{cc}1&0\\\\1&1\end{array}\right)$ and $y = \left(\begin{array}{cc}n&0\\\\0&1\end{array}\right).$

$\endgroup$
2
  • 2
    $\begingroup$ The subgroup generated by $x$ is not normal! $\endgroup$
    – Steve D
    May 15, 2011 at 20:08
  • 1
    $\begingroup$ But the normal subgroup generated by $x$ is normal and infinitely generated. It is isomorphic to the additive group of diadic rational numbers. $\endgroup$
    – user6976
    May 16, 2011 at 5:59
9
$\begingroup$

There are finitely presented metabelian groups containing the lamplighter groups. One of them was constructed by Baumslag: $\langle a,b,c \mid a^2=1, [b,c]=1, [a^b,a]=1, a^c=a^ba\rangle$.

$\endgroup$
0
2
$\begingroup$

By the way, you may enjoy the fact, due to G. Baumslag, that a standard wreath product $W\wr G$ with $W\neq 1$ and $G$ infinite, is never finitely presented; see Gilbert Baumslag. Wreath products of finitely presented groups. Math. Z. 75 , 22-28, 1961. For finite presentability of permutational wreath products, see a paper by Cornulier: http://www.normalesup.org/~cornulier/wrea_fp.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.