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.