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

Antonio Alfieri
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2 Answers2

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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).

hmakholm left over Monica
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  • 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? – Timm von Puttkamer Nov 02 '14 at 17:49
  • @Jim: By "all permutations" I mean all permutations. It appears you're missing those that have infinite support. – hmakholm left over Monica Nov 02 '14 at 20:14
  • 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. – Timm von Puttkamer Nov 02 '14 at 21:40
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    @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. – hmakholm left over Monica Nov 02 '14 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 $x \mapsto \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}$.

Martin Brandenburg
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