Question

If G is the additive group $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$, what are the elements of the subgroup G(2)? Of G(p) for any positive prime p?

Short answer

The elements in the subgroup G(2) is $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},2\cancel{|}\mathrm{a}\right\}$ and $\mathrm{G}\left(\mathrm{p}\right)=\left\{\left[\frac{\mathrm{a}}{{\mathrm{p}}^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},\mathrm{p}\cancel{|}\mathrm{a}\right\}$ .

Step-by-step solution

Elements of subgroup G(2)

Let G be the additive group $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$then the elements in the subgroup G(2) will be $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{\mathrm{b}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\exists \mathrm{n}\in \mathrm{N}:\left[\frac{2^{\mathrm{n}}\mathrm{a}}{\mathrm{b}}\right]=\left[0\right]\right\}$.

Here, $\left[\frac{a}{b}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$if and only if there exists some $\mathrm{n}\in \mathrm{N}$such that $b|2^{n}a$.

If (a,b)=1 then there exists $n\in N$such that $b=2^n$.

Therefore, G(2) can be written as $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},2\cancel{|}\mathrm{a}\right\}$.

The subgroup G(p)  

The subgroup G(2) is $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},2\cancel{|}\mathrm{a}\right\}$. Similarly, for any positive prime p, the subgroup G(p) will be role="math" localid="1659615113258" $\mathrm{G}\left(\mathrm{p}\right)=\left\{\left[\frac{\mathrm{a}}{{\mathrm{p}}^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},\mathrm{p}\cancel{|}\mathrm{a}\right\}$.

Therefore, $\mathrm{G}\left(\mathrm{p}\right)=\left\{\left[\frac{\mathrm{a}}{{\mathrm{p}}^{\mathrm{n}}}\right]\in \mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}|\mathrm{n}\in \mathrm{N},\mathrm{p}\cancel{|}\mathrm{a}\right\}$.