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I'm considering the possible combinations of these properties for an isometry group: "finite", "finitely generated", "discrete".

Obviously, a finite group is necessarily finitely generated and discrete. The remaining possibilities are for infinite groups.

The group may be both finitely generated and discrete. Example: The group of translations in $\mathbb R$ by integers, is generated by a single translation ($x\mapsto x+1$).

The group may be finitely generated but not discrete. Example: The group of rotations in $\mathbb R^2$ by multiples of an irrational angle (like $1$ radian $= 1/(2\pi)$ cycles), maps a single point to a set of points on a circle which come arbitrarily close to the original point.

The group may be not discrete nor finitely generated. Example: The group of all translations in $\mathbb R$.

Can the group be discrete but not finitely generated? If we use $\mathbb R^\infty$ (the space of sequences with finitely many non-zero terms), the group generated by reflections along the basis vectors $\{(1,0,0,0,\cdots),(0,1,0,0,\cdots),(0,0,1,0,\cdots),\cdots\}$ seems to work. But I don't want to use an infinite-dimensional space.

I think I've proven that a group of finite-dimensional translations, if discrete, must be finitely generated. The proof also works for rotations in $\mathbb R^2$. I could try to generalize it to rotations in higher dimensions (of Euclidean space), but there's still hyperbolic space.

I found this about Fuchsian groups, which are discrete isometry groups for the hyperbolic plane:

There are some variations of the definition: sometimes the Fuchsian group is assumed to be finitely generated...

Is that a necessary assumption? Are there Fuchsian groups that aren't finitely generated?

mr_e_man
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It is easy to find a discrete group of isometries of the hyperbolic plane which is isomorphic to $F_2$, the free group with two generators. The generators could be e.g. "go 10 units" and "rotate 90 degrees right, go 10 units, rotate 90 degrees left". (You can see an interactive demo of this here -- press 'g' to mark places, the distance here is smaller than 10 units, but large enough that you will never reach the starting point unless by retracing your path, thus proving that the group is indeed free.)

And $F_2$ has subgroups which are not finitely generated. Subgroups of finitely generated groups are not necessarily finitely generated

Zeno Rogue
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    This looks good. I read those posts before, so I was expecting something like this. I don't yet see why it's isomorphic to $F_2$, but it's probably obvious... – mr_e_man Oct 07 '18 at 01:01
  • I have added some extra explanation. – Zeno Rogue Oct 07 '18 at 01:11
  • Sorry, that demo doesn't work in my browser. I have played a similar game http://www.madore.org/~david/math/hyperbolic-maze-2.html , so I get the basic idea. I would still like to prove it algebraically. – mr_e_man Oct 07 '18 at 01:30
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    Maybe the browser is too old? It is rather clear on the picture, but an algebraic proof can be given too :) – Zeno Rogue Oct 07 '18 at 02:16
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    ...So, a point's orbit under this group would look something like this? https://en.wikipedia.org/wiki/Order-4_apeirogonal_tiling That does look like the free group. – mr_e_man Oct 07 '18 at 03:33
  • Okay, I took two orthogonal translations $a$ and $b$, both by distance $\phi$ satisfying $\cosh\phi=3$. I found that the composition $aba^{-1}b^{-1}ab\cdots$ ($n$ factors) sends the origin to a point $d$ distance away, with $$\cosh d=1+2n^2.$$ This shows several things. With $n=4$, we get $d\neq0$, which shows that $aba^{-1}b^{-1}$ is not the identity, so $ab\neq ba$. With $n$ increasing, the distance $d$ always increases (doesn't go near $0$), suggesting that the group is discrete. I haven't looked at other elements of the group (except $a^n$ and $b^n$). But this is good enough for now. :) – mr_e_man Oct 07 '18 at 10:36
  • It appears that any free group $F_n$ (with $0<n<\infty$) can be represented in the hyperbolic plane, with the vertices of the apeirogonal tiling ${\infty,2n}$. – mr_e_man Oct 07 '18 at 11:23
  • Yes, this is the correct picture (I have found it too, but I did not like how it was not vertex-centered). I assume your $\phi$ is the edge length of ${\infty,4}$? This is an elegant result, but you would have to check other elements for a full proof. – Zeno Rogue Oct 07 '18 at 15:51
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    As a computer scientist, I prefer proofs based on asymptotic analysis rather than exact computations: it should be easy to show inductively that by composing $n$ elements of form "move 10 units" the total distance is at least $9n$ units, and the angle between $OX$ and your next move is at least 80 degrees (numbers based on a guess, maybe you should change them). Though apeirogons whose vertices lie on horocycles are more elegant than ones where they lie on equidistant curves :) – Zeno Rogue Oct 07 '18 at 15:56
  • Applying the hyperbolic law of cosines to a triangle formed by the origin, its image by a transformation, and its image by the transformation composed with $a$, we get $$\cosh d'=\cosh d\cosh\phi-\sinh d\sinh\phi\cos\theta=3\cosh d-2\sqrt2\sinh d\cos\theta\overset{?}{<}\cosh d$$ $$1/\sqrt2\overset{?}{<}\tanh d\cos\theta$$ And the transformation composed with $a^{-1}$ or $b$ or $b^{-1}$ would have $\theta$ replaced with $\theta+n\frac\pi2$, or $\cos$ replaced with $\pm\cos$ or $\pm\sin$. This results in 4 inequalities, of which only 1 can be satisfied. (...) – mr_e_man Oct 17 '18 at 08:24
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    (...) (For example, with both $\cos$ and $\sin$, square the inequalities and add them to get $1/2+1/2<\tanh^2 d(\cos^2\theta+\sin^2\theta)$; that is, $1<\tanh^2 d$, a contradiction.) This shows that, for any transformation, only one of the four compositions with $a,b,a^{-1},b^{-1}$ may decrease the distance from the origin. The others must increase the distance. It's easy to complete the proof from here. – mr_e_man Oct 17 '18 at 08:32