How is $ F(2,m) = \langle a_1, a_2,\ldots, a_m \mid a_i a_{i+1} = a_{i+2} \rangle $ a group?

Question

A Fibonacci group is defined as the group with the presentation $$ F(2,m) = \langle a_1, a_2,\ldots, a_m \mid a_i a_{i+1} = a_{i+2} \rangle $$ where the indexes are reduced modulo $m$.

Can someone describe how this a group ? Please post some other references.

Where do you have this concept from? (and what does that notation even mean?) — hmakholm left over Monica, Sep 25 '19 at 20:39
https://www.encyclopediaofmath.org/index.php/Fibonacci_group — user157835, Sep 25 '19 at 20:43
Consider the free group $G$ on generators $a_1,\ldots, a_n$. In it, let $N$ be the smallest normal subgroup $N$ that contains all $a_ia_{i+1}a_{i+2}^{-1}$. Then $F(2,m)=G/N$. — Hagen von Eitzen, Sep 25 '19 at 20:44
@user157835: You were missing the angle brackets, and it would be nice to have stated explicitly that this was supposed to be a group presentation. — hmakholm left over Monica, Sep 25 '19 at 20:47
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Shaun · Answer 1 · 2019-09-26T09:05:36.440

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Let $F$ be the free group on $\{a_i\}_{i=1}^m$. Quotient out by the normal subgroup $N$ generated by the relations $a_ia_{i+1}=a_{i+2}$ (i.e., the elements (or "relators") $a_ia_{i+1}a_{i+2}^{-1}$), taking the subscripts modulo $m$. The resulting group $G\cong F/N$ is the group defined by your group presentation.

Such a description is typical (and in fact a definition of) a group presentation.

For a reference, you can't get more comprehensive than this survey of Fibonacci groups.

edited Sep 26 '19 at 09:05

answered Sep 26 '19 at 02:11

Shaun

44,997

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Is it clear what "the normal subgroup generated by the relations $a_ia_{i+1}=a_{i+2}$" means? It is actually the normal subgroup generated by the elements $a_ia_{i+1}a_{i+2}^{-1}$. – Derek Holt Sep 26 '19 at 08:29
Thank you, @DerekHolt. I've edited it now. – Shaun Sep 26 '19 at 09:06

How is $ F(2,m) = \langle a_1, a_2,\ldots, a_m \mid a_i a_{i+1} = a_{i+2} \rangle $ a group?

1 Answers1