Question

Show that ${\text{Q}}^{\text{**}}$, the multiplicative group of positive rational numbers, is not a cyclic group. [Hint: if $\mathbf{1}\ne \text{r}\in {\text{Q}}^{\text{**}}$, then there must be a rational between r and r².]

Short answer

Answer

It is proved that ${\text{Q}}^{\text{**}}$is not a cyclic group.

Step-by-step solution

Step-by-Step Solution Step 1: Condition for a group to be cyclic

A group G is considered cyclic if it can be written in the form of $\langle \text{r}\rangle \mathbf{\text{=}}\left\{{\text{r}}^{\text{n}}\left|\text{r}\in \mathbb{Z}\right.\right\}$ and the element is called the generator of the group.

Show that Q** is not a cyclic group

Suppose $Q^{\ast \ast }$is cyclic and generated by element a. So, it must be $r\ne 1$ since 1ⁿ = 1 for all $n\in \mathbb{Z}$.

It can be assumed that r > 1 without the loss of generality. Take the inverse r^-1 as the generator. Note that the function $\varphi :Z\to Q^{\ast \ast }$with $\varphi n=r^n$is monotone, which means that m < n and r^m < rⁿ.

To check, take $r=\frac{a}{b}$with a > b by the reasoning. Then,

$\begin{array}{c}{\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}\\ =\frac{a^m{b}^{n-m}}{b^n}\\ <\frac{a^n}{b^n}\\ ={\left(\frac{a}{b}\right)}^n\end{array}$

Consider that $\frac{r+r^2}{2}\in Q^{\ast \ast }$and $r<\frac{r+r^2}{2}<r^2$, so there cannot be any $m\in Z$such that $r^m=\frac{r+r^2}{2}$, which contradicts r generating $Q^{\ast \ast }$.

Thus, it is proved that $Q^{\ast \ast }$is not a cyclic group.