In my abstract algebra course, I learned about finitely generated groups. One of the exercises proves that a subgroup of a finitely generated group is not necessarily finitely generated itself.
The exercises looks at the group $S\mathbb{Z}, \circ$, the group of permutations of $\mathbb{Z}$. Two permutations are defined: $$\sigma: \mathbb{Z} \to \mathbb{Z}: z \to z +1,$$ (so right-shift) and $\tau_{i,j}$ which fixes all integers, except for $i,j$ which are swapped.
We then considered $G = \text{grp}\{\sigma, \tau_{1,2}\}$ (=group generated by...) and need to prove that $\tau_{i,j} \in G$. I remarked that:
$\tau_{n, n+1} = \sigma^{n-1} \circ \tau_{1,2} \circ \sigma^{n-1}$
$\tau_{i,j} = \tau_{j-1, j} \circ \cdot \circ \tau_{i, i+1} \circ \cdots \tau_{j-1, j}$.
and this shows that $\tau_{i,j} \in G$.
Now define $H = \text{grp}(\tau_{i,j} \text{ for all } i,j \in \mathbb{Z} \text{ with } i > j\}$. This is a subgroup of $G$ which is not finitely generated.
questions:
- is there a more direct way to prove that $\tau_{i,j} \in G$?
- the show that $H$ is not finitely generated, I thought to pick a finite number of $\tau_{i,j}$. Then there is a largest integer $k$ that is swapped by these generators. This shows that $\tau_{k+1, k+2}$ would not be contained in this finitely generated group, so $H$ can not be finitely generated. Is this correct? Or is it possible that $H$ has other generators than transpositions?
For $2$ I would first prove that every element of $H$ moves a finite number of elements (which is not that hard since it is a finite product of transpositions). It follows that a finite number of elements of $H$ can only move a finite number of elements of $\mathbb Z$ (the union of the elements that are moved by each of the permutations). It follows that if $i,j$ are not in this set then $\tau_{i,j}$ is not generated by these elements.