A non-cyclic proper subgroup of $(\Bbb{Q},+)$

Question

I need to show that $H$ is a proper non-cyclic subgroup of $(\Bbb{Q}, +)$ where

$$H = \left\{\frac{m}{2^n}: m \in \Bbb{Z}, n\in\Bbb{N}\right\} $$

First if I need to show it is a subgroup, I need to show that it is non-empty subset of $\Bbb{Q}$ and $a + b^{-1} \in H$ for all $a, b \in H$.

This is clear that $H$ is a non-empty subset of $\Bbb{Q}$.

Now let $\frac{m_1}{2^{n_1}}$ and $\frac{m_2}{2^{n_2}}$ be in $H$.

Then it is clear that $\frac{m_1}{2^{n_1}} - \frac{m_2}{2^{n_2}}$ will be in $H$.

It is clear that $H$ will be proper.

Next, how can I show that it is non-cyclic?

Answer 1

To be cyclic is to be generated by one element.

A typical non-identity element is a number of the form $\dfrac m {2^n}$ for some $m\in\mathbb Z, n \in \{0,1,2,3,\ldots\}.$

The subgroup generated by $\dfrac m {2^n}$ is $\left\{ \dfrac{mk}{2^n} : k\in \mathbb Z \right\}.$ The number $\dfrac 1 {2^{n+1}}$ is a member of $H$ but not of the group generated by $\dfrac m {2^n}.$

Answer 2

You have to show that $a-b\in H$, not $a+b^{-1}$, because you're considering $\mathbb{Q}$ with respect to addition.

Saying “it is clear” is not really a proof.

Consider instead $$ \frac{x}{2^m}-\frac{y}{2^n}=\frac{2^nx-2^my}{2^{m+n}} $$ which actually shows that the difference belongs to $H$.

Why is $H$ proper? Because there are no integers $m$ and $n\ge0$ such that $$ \frac{1}{3}=\frac{m}{2^n} $$ because $3$ does not divide $2^n$.

Finally, take $m/2^n\in H$ and find something in $H$ that's not an integral multiple of this element. Hint: use a larger denominator.