Question

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

Answer 1

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:

`[g, h] = 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 `[a^n, b^n]`.

Any element of `S` can be written as a product of elements, each of which is of the form `[a^n, b^n]` or `[b^n, a^n] = [a^n, b^n]^(-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 `[a^n, b^n]` and/or `[b^n, a^n]`.

• 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 `[a^n, b^n][b^n, a^n]` or `[b^n, a^n][a^n b^n]` to reduce to the minimum representation.

Hence we find that `S` is essentially the free group generated by `[a^n, b^n]` for `n = 1, 2, 3,...`. Hence it is not finitely generated.