Can you decide whether the commutator subgroup of a f.p. group is f.g? Is the following algorithmic problem known to be decidable/undecidable?
Input: a finite group presentation $P$.
Decide: is the commutator subgroup of the group presented by $P$ finitely generated?
 A: It's undecidable.
Lemma: a group $G$ is nontrivial if and only if the free product $H=G\ast\mathbf{Z}$ has an infinitely generated derived subgroup.
Proof: assume $G$ finitely generated and nontrivial. The kernel $N$ of the canonical epimorphism $H\to\mathbf{Z}$ is isomorphic to $G^{\ast\mathbf{Z}}$ and hence is infinitely generated. Since $H/[H,H]$ is finitely generated abelian, so is $N/[H,H]$; if $[H,H]$ were finitely generated, so would be $N$, a contradiction. Hence $[H,H]$ is infinitely generated. In case $G$ is infinitely generated and if by contradiction $[H,H]$ is finitely generated, then its generators belong to $G_1\ast\mathbf{Z}$ for some proper subgroup $G_1$ of $G$, which is obviously a contradiction.$\Box$
The result then from the fact that if we have a Turing machine $X$ whose input is a finite presentation $P$ and whose answer is yes or no according to whether the group presented by $P$ has a finitely generated derived subgroup, then if we input in $X$ the presentation obtained from $P$ by adding a generator, the output is yes or no according to whether the group presented by $P$ is trivial. Thus the resulting machine solves the triviality problem, and it is known that there is no such machine. 
