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What groups can arise as being generated by two rotations in $\mathbb R^2$ by angles $\not \in \mathbb Q\pi$?

If the centers of the rotations coincide, then the rotations commute and generate some quotient of $\mathbb Z^2$ that can be understood by the number theoretic properties of the two rotation angles.

My guess would be that the group generated is always nonabelian free when the centers are distinct, but I don't have a proof. It seems difficult to understand in general because the group is indiscrete, so e.g. it might be hard to set up a ping pong argument to prove freeness.

NB that the relative position of the centers doesn't matter because you can always rotate and scale the picture. It's only a question of whether the centers are identical or not identical.

This came out of something I was thinking about in my research, but it's also related to this fascinating paper I just learned about

Edit Many pointed out that the group can never be free for a trivial reason: commutators of rotations give translations, so these groups will have plenty of commuting elements. In other words, these groups are metabelian. The next question would be whether the group generated is a free metabelian group on two generators. Or perhaps we ask whether the monoid generated is free (thereby forbidding commutators).

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    $\begingroup$ It’s a subgroup of $\mathbb R^2\rtimes SO(2)$ which is metabelian, so it’s certainly not non-abelian free. $\endgroup$
    – Corentin B
    Jan 11 at 22:31
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    $\begingroup$ The group of all rigid motions of the plane is an extension of the rotation group by the translations, so is soluble. Any subgroup of a soluble group is soluble, whereas the free group on two generators is not soluble. $\endgroup$ Jan 11 at 22:40
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    $\begingroup$ When you consider the subgroup generated by a “generic” rotation and a translation, you get a subgroup isomorphic to the lamplighter group $\mathbb Z\wr\mathbb Z$. In general, for a rotation and a translation, we get a quotient of $\mathbb Z\wr\mathbb Z$. This is the kind of groups one should expect (even tho, for two rotations, the extension is not longer split). $\endgroup$
    – Corentin B
    Jan 11 at 22:41
  • $\begingroup$ However the group of motions of the plane contains a free subsemigroup on 2 generators, so one can ask when 2 rotation generate such a subsemigroup. $\endgroup$
    – YCor
    Jan 11 at 22:49
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    $\begingroup$ @CorentinB indeed two generic elements probably generate a free metabelian group on 2 generators. $\endgroup$
    – YCor
    Jan 11 at 22:56

2 Answers 2

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The commutator of any two elements of your group is a translation, so they all commute. So for example if $a$ and $b$ are two elements of your group then $[a,b]$ commutes with $[a^b,b]$. This is a relation that does not hold in the free group.

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    $\begingroup$ Great observation, I should have thought a bit harder. Thank you Dave. The next question would be if the generated group can be a free metabelian group on two generators as @YCor suggested in the comments. $\endgroup$ Jan 12 at 4:05
  • $\begingroup$ @EthanDlugie I more or less had an argument in mind. So I have eventually posted an answer mathoverflow.net/a/462050/14094 $\endgroup$
    – YCor
    Jan 12 at 12:29
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A generic (i.e., outside a countable union of Zariski-closed proper subsets) $k$-tuple in $G=\mathrm{SO}(2)\ltimes\mathbf{R}^2$ freely generated a free metabelian group. In particular, it freely generates a free semigroup.

Proof. Let $w$ belong to $F_k/F_k''-\{1\}$. Then the set $X_w\subset G^k$ of $k$-tuples in $G$ with $w(g_1,\dots,g_k)=0$ is a Zariski-closed subset. To conclude, it is enough to prove that $X_w\neq G^k$.

Otherwise, since $G$ contains a copy of $H=\mathbf{Z}\wr\mathbf{Z}$, we deduce that $w$ vanishes on $H^k$. But $H$ embeds as a Zariski-dense subgroup in the affine group $A=\mathbf{C}\rtimes \mathbf{C}^*$, so $w$ also vanishes on $A$. But finally, the Magnus embedding says that finitely generated free metabelian groups embed in $A$, and we get a contradiction, since $w$ doesn't vanish in the free metabelian group $F_k/F_k''$.

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    $\begingroup$ Here is a small variation that avoid to pass by $A$. The group $G$ also contains a copy of $\mathbb Z^k\wr \mathbb Z^k$ (generated by $k$ generic translations and $k$ generic rotations with same center) which in turn contains $F_k/F''_k$ by some other version of the Magnus embedding. $\endgroup$
    – Corentin B
    Jan 12 at 13:26
  • $\begingroup$ Is the wreath product isomorphic with the free metabelian group? $\endgroup$ Jan 12 at 15:46
  • $\begingroup$ @EthanDlugie No, it isn't. The free metabelian group is not a semidirect product of 2 abelian groups. $\endgroup$
    – YCor
    Jan 12 at 16:24

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