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In the group of isometries of the plane, let $r$ be a rotation by $\frac{2\pi}5$ radians and let $s$ be a translation. I'd like to find a finite presentation of the subgroup $G=\langle r,s\rangle$. Some obvious relations I can think of are: $$r^5=1\\(sr)^5=1\\srsr^{-1}=rsr^{-1}s$$ but these seem insufficient, because when I put them into a Knuth-Bendix calculator, it runs forever and generates larger and larger rewriting systems that don't reduce $(s^2r)^5$ to the identity. I know Knuth-Bendix can fail, but I sort of expect it to work here because I read that Euclidean groups are automatic.

How can we find (with proof) a presentation of $G$? How can we even see whether $G$ is finitely presentable?

(I could find a matrix representation like in this answer, but I'm not sure how that would help me find a presentation.)

Edit: I added the relation $sr^2sr^{-2}=r^2sr^{-2}s$ based on some geometric reasoning (tiling a pentagon of side length 2 with shapes of side length 1 represented by the relations). KB still runs forever, but now it eventually reaches rewriting systems that establish $(s^nr)^5=1$ for $n=2,3,4$, so this seems like a more promising candidate for the correct presentation.

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    $\begingroup$ The group is virtually abelian, and therefore automatic, but that does not imply that Knuth-Bendix will work. $\endgroup$ Commented Jun 6 at 11:25

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Here is a Magma calculation that shows that your presentation with the extra relation defines a group that is an extension of a free abelian group of rank 4 (the normal closure of $\langle s \rangle$ in $G$) with the cyclic group $\langle r \rangle$ of order $5$. Note that $(g,h)$ in Magma denotes the commutator $g^{-1} h^{-1}gh$.

> G<r,s>:= Group< r,s | r^5, (s*r)^5, (s,r*s*r^-1), (s,r^2*s*r^-2) >; > N := ncl< G | s >; > Index(G,N); 5 > Rewrite(G,~N); > N; Finitely presented group N on 4 generators Index in group G is 5 Generators as words in group G N.1 = s N.2 = r * s * r^-1 N.3 = s^r N.4 = r^2 * s * r^-2 Relations (N.1, N.2) = Id(N) (N.1, N.4) = Id(N) (N.2, N.4) = Id(N) (N.3, N.1) = Id(N) (N.3, N.2) = Id(N) (N.4, N.3) = Id(N)

If you do the same calculation with the final relator $(s,r^2sr^{-2})$ omitted, then the normal closure $N$ os $\langle s \rangle$ in $G$ is a group with $4$ generators and $5$ relators, so it cannot be free abelian of rank $4$.

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  • $\begingroup$ Thanks! I follow why the normal closure of $\langle s\rangle$ consists of all the translations and is therefore an abelian normal subgroup with index 5 generated by at most 4 elements. How do we know it is free abelian of rank exactly 4? (Does it include arbitrarily small translations?) $\endgroup$ Commented Jun 6 at 19:52
  • $\begingroup$ The presentation that was computed for $N$ has four generators and $6$ relations that say that each pair of these generators commutes. This presentation defines a free abelian group of rank $4$. $\endgroup$ Commented Jun 6 at 19:59
  • $\begingroup$ Thanks - I see that about the computed group and am just lacking some understanding of my desired group. Is it the case for every odd $n$ that the subgroup of $(\Bbb C,+)$ generated by the primitive $n$th roots of unity is free abelian of rank $n-1$? I'll post a separate question. $\endgroup$ Commented Jun 6 at 22:01
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    $\begingroup$ @Karl I think the rank is given by Euler's phi function. So it is $n-1$ when $n$ is prime. I think you can see this using cyclotomic polynomials. $\endgroup$ Commented Jun 19 at 10:19

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