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

1 Answer
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:

[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.