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Is there a group with trivial abelianization having as subgroup a group isomorphic to $\mathbb{Z} \times \mathbb{Z}$?

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    $\begingroup$ "infinite" is redundant here; a group containing $\mathbb{Z} \times \mathbb{Z}$ is infinite. Good question by the way. I am quite sure that there are examples. $\endgroup$ Nov 1, 2014 at 21:48
  • $\begingroup$ Oh, yes I'm sorry! $\endgroup$ Nov 1, 2014 at 21:50

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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).

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  • $\begingroup$ Do you mean by the group of all permutations those permutations that have finite support? If this is the case, then $\mathbb Z \times \mathbb Z$ does not include into the group. What am I missing? $\endgroup$ Nov 2, 2014 at 17:49
  • $\begingroup$ @Jim: By "all permutations" I mean all permutations. It appears you're missing those that have infinite support. $\endgroup$ Nov 2, 2014 at 20:14
  • $\begingroup$ I see. But then I can't follow your argument in the second paragraph. Any composition of two bijections of finite support has again finite support. $\endgroup$ Nov 2, 2014 at 21:40
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    $\begingroup$ @Jim: I'm not speaking of bijections with finite support at all. I'm speaking of ones where the complement of the support is infinite. That's possible even if the support itself is also infinite -- for example for the bijection that adds $2$ to every odd number and leaves the even numbers unchanged. $\endgroup$ Nov 2, 2014 at 21:55
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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 $x \mapsto \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}$.

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  • $\begingroup$ Does $\mathbb{Z} \times \mathbb{Z}$ embed into $\mathrm{SL}_2(\mathbb{C})$? $\endgroup$ Nov 1, 2014 at 22:51
  • $\begingroup$ x @Martin: Yes, by $(a,b)\mapsto \mathrm{Diag}(e^{a+bi},e^{-a-bi})$, thanks to $\pi$ being irrational. $\endgroup$ Nov 1, 2014 at 23:15
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    $\begingroup$ More generally $\mathbb Z^n$ embeds into $\mathrm{SL}_2(\mathbb Q)$ (or $\mathbb Q^n$ into $\mathrm{SL}_2(\mathbb R)$) by $(x_1,x_2,\ldots,x_n)\mapsto \mathrm{Diag}(2^{x_1}3^{x_n}\cdots p_n^{x_n}, 2^{-x_1}3^{-x_2}\cdots p_n^{-x_n})$ $\endgroup$ Nov 1, 2014 at 23:51

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