## Can we have a finite cyclic group of rational numbers (under multiplication)? [closed]

I am a PDE guy, who works in imaging. We are trying to exploit the inherent group structure of an image, i.e. we consider a representation of the similitude group ($SIM(2)=\mathbb{R}^2\ltimes(SO(2)\times\mathbb{R}^{+})$) (group of scaling, translation and rotation) on the space $\mathbb{L}_{2}(\mathbb{R}^2)$, which is the space of images.

We work with a generalization of the wavelet transform on the $SIM(2)$ group. We have been trying to discretize the group of scalings or dilations, $a\in\mathbb{R}^{+}$. Now in our context we wish to stay away from "a=0" (for numerical as well as theoretical purposes) and have a maximum scaling value ($a<\infty$, practical purposes). So we are trying to come up with a discrete group over $\mathbb{R}^{+}$ or $\mathbb{R}$ (where we substitute $\tau=log(a)$). We were thinking along the lines of either a cyclic multiplicative modulo group of rationals (between say $\frac{1}{d}$ and $b$) or a cyclic additive modulo group of reals (between say $-d$ and $b$) respectively where $d,b$ are predefined integers or real numbers.

Though the whole construction of substituting a infinite group with a a finite group (with a different operation) may seem dubious, it allows us to justify the numerics in best possible (mathematical) way.

So my question is such a construction of a finite cyclic group possible.

It might be the case that due to my lack of knowledge in the field of groups this question may have an obvious answer. Nevertheless I would be grateful if someone could guide me in the right direction.

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The only finite subgroup of the additive group $\mathbf{R}$ is the trivial group, and the only finite subgroups of the multiplicative group $\mathbf{R}^\times$ are the trivial group and the order-$2$ group generated by $-1$. – Chandan Singh Dalawat May 30 at 8:00

## closed as off topic by Dan Petersen, Chandan Singh Dalawat, Chris Godsil, Felipe Voloch, Andreas BlassMay 30 at 13:27

Since you want a multiplicative group of rationals bounded between $\frac{1}{d}$ and $b$, you can prove its non-existence in an easy way: assume your group exists and is non-trivial and let $g\neq1$ be one of its elements. Consider $g^n$. If $0 < g < 1$, then $g^n\to0$; if $g>1$, then $g^n\to\infty$. On the other hand, $g^n$ must belong to your group for all $n$, showing that such bounds you want cannot exist. Analogue procedure shows that a bounded additive group of real numbers cannot exist.