// Workers AI · dad joke modeWhat did the Fibonacci group say? We're a growing concern.

In mathematics, for a natural number $n\geq 2$, the nth Fibonacci group, denoted $F(2,n)$ or sometimes $F(n)$, is defined by n generators $a_{1},a_{2},\dots ,a_{n}$ and n relations:

• $a_{1}a_{2}=a_{3},$
• $a_{2}a_{3}=a_{4},$
• $\dots$
• $a_{n-2}a_{n-1}=a_{n},$
• $a_{n-1}a_{n}=a_{1},$
• $a_{n}a_{1}=a_{2}$.

These groups were introduced by John Conway in 1965.

The group $F(2,n)$ is of finite order for $n=2,3,4,5,7$ (order 1,8,5,11,29, respectively), and infinite order for $n=6$ and $n\geq 8$. The infinitude of $F(2,9)$ was proved by computer in 1990.

Kaplansky's unit conjecture

See also: Kaplansky's conjectures

From a group $G$ and a field $K$ (or more generally a ring), the group ring $K[G]$ is defined as the set of all finite formal $K$-linear combinations of elements of $G$ − that is, an element $a$ of $K[G]$ is of the form $a=\sum _{g\in G}\lambda _{g}g$, where $\lambda _{g}=0$ for all but finitely many $g\in G$ so that the linear combination is finite. The (size of the) support of an element $a=\sum \nolimits _{g}\lambda _{g}g$ in $K[G]$, denoted $|\operatorname {supp} a\,|$, is the number of elements $g\in G$ such that $\lambda _{g}\neq 0$, i.e. the number of terms in the linear combination. The ring structure of $K[G]$ is the "obvious" one: the linear combinations are added "component-wise", i.e. as $\sum \nolimits _{g}\lambda _{g}g+\sum \nolimits _{g}\mu _{g}g=\sum \nolimits _{g}(\lambda _{g}\!+\!\mu _{g})g$, whose support is also finite, and multiplication is defined by $\left(\sum \nolimits _{g}\lambda _{g}g\right)\!\!\left(\sum \nolimits _{h}\mu _{h}h\right)=\sum \nolimits _{g,h}\lambda _{g}\mu _{h}\,gh$, whose support is again finite, and which can be written in the form $\sum _{x\in G}\nu _{x}x$ as $\sum _{x\in G}{\Bigg (}\sum _{g,h\in G \atop gh=x}\lambda _{g}\mu _{h}\!{\Bigg )}x$.

Kaplansky's unit conjecture states that given a field $K$ and a torsion-free group $G$ (a group in which all non-identity elements have infinite order), the group ring $K[G]$ does not contain any non-trivial units – that is, if $ab=1$ in $K[G]$ then $a=kg$ for some $k\in K$ and $g\in G$. Giles Gardam disproved this conjecture in February 2021 by providing a counterexample. He took $K=\mathbb {F} _{2}$, the finite field with two elements, and he took $G$ to be the 6th Fibonacci group $F(2,6)$. The non-trivial unit $\alpha \in \mathbb {F} _{2}[F(2,6)]$ he discovered has $|\operatorname {supp} \alpha \,|=|\operatorname {supp} \alpha ^{-1}|=21$.

The 6th Fibonacci group $F(2,6)$ has also been variously referred to as the Hantzsche-Wendt group, the Passman group, and the Promislow group.

External links

• An alternative proof that the Fibonacci group F(2,9) is infinite by Derek K. Holt (PostScript file).
• "Fibonacci group". Encyclopedia of Mathematics.