# Is it possible for a finitely-generated group to contain subgroups that are not finitely-generated ? True or False. Prove your conclusion.

Feb 20, 2018

It is possible.

#### Explanation:

The classic example is the commutator subgroup of the free group on two generators.

Let:

$G = < a , b >$

If $g , h \in G$ then the commutator of $g$ and $h$ is:

$\left[g , h\right] = {g}^{- 1} {h}^{- 1} g h$

The subgroup of $G$ generated by its commutators is not finitely generated, but I have not encountered a simple proof.

We can make things simpler by specifying the subgroup $S$ generated by all commutators of the form $\left[{a}^{n} , {b}^{n}\right]$.

Any element of $S$ can be written as a product of elements, each of which is of the form $\left[{a}^{n} , {b}^{n}\right]$ or $\left[{b}^{n} , {a}^{n}\right] = {\left[{a}^{n} , {b}^{n}\right]}^{- 1}$. Moreover, the minimum length representation in that form is unique.

Given a product of $a$'s, ${a}^{- 1}$'s, $b$'s and ${b}^{- 1}$'s in $S$, proceed as follows:

• Strip out any element-inverse element pairs to reduce the product to minimum form, i.e. get rid of combinations like $a {a}^{- 1}$ or ${b}^{- 1} b$.

• The cleaned up form will start with a block of ${a}^{- 1}$'s or ${b}^{- 1}$'s. The length of this block allows you to identify the first commutator in a representation as a product of commutators of the form $\left[{a}^{n} , {b}^{n}\right]$ and/or $\left[{b}^{n} , {a}^{n}\right]$.

• The end of the commutator may require some element-inverse element pairs to be added in order to reconstitute it. Add just as many as necessary before splitting off the first commutator.

• Repeat with the remainder of the product to recover the remaining commutators.

• Once all commutators have been recovered, strip out any adjacent pairs of commutators of the form $\left[{a}^{n} , {b}^{n}\right] \left[{b}^{n} , {a}^{n}\right]$ or $\left[{b}^{n} , {a}^{n}\right] \left[{a}^{n} {b}^{n}\right]$ to reduce to the minimum representation.

Hence we find that $S$ is essentially the free group generated by $\left[{a}^{n} , {b}^{n}\right]$ for $n = 1 , 2 , 3 , \ldots$. Hence it is not finitely generated.