A group with trivial abelianization and a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$ Is there a group with trivial abelianization having as subgroup a group isomorphic to $\mathbb{Z} \times \mathbb{Z}$? 
 A: Try the group of all permutations of $\mathbb Z\times \mathbb Z$ (as a set).
It is its own commutator subgroup (i.e., it has trivial abelianization): Every permutation that leaves infintely many numbers alone is a commutator, and you ought to be able to construct anything you want as a product of two of those.
Also, it contains $\mathbb Z\times \mathbb Z$ as a subgroup, by its action on itself.
(This argument easily generalizes to show that every group is isomorphic to a subgroup of something with trivial abelianization).
A: Yes. If $k$ is an infinite field and $n \geq 2$, then $\mathrm{SL}_n(k)$ is known to be an infinite perfect group. Now $\mathbb{Z} \times \mathbb{Z}$ embeds into $\mathrm{GL}_2(\mathbb{C})$ via diagonal matrices, because $\mathbb{Z}$ embeds into $\mathbb{C}^*$ (rotation with irrational angle). But $\mathrm{GL}_2(\mathbb{C})$ embeds into $\mathrm{SL}_3(\mathbb{C})$ via $A \mapsto \begin{pmatrix} A & 0 \\ 0 & \det(A)^{-1} \end{pmatrix}$.
PS: As mentioned in the comments, a much more easier embedding is the following: $\mathbb{Z} \times \mathbb{Z}$ (in fact every $\mathbb{Z}^n$) embeds into $\mathbb{Q}^*$ via $(a,b) \mapsto 2^a 3^b$. And $\mathbb{Q}^*$ embeds into $\mathrm{SL}_2(\mathbb{Q})$ via $x \mapsto \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}$.
