Can the infinite cyclic group be understood as the group of all rational numbers under addition?

Question

I've just started the beginnings of abstract algebra and came across the concept of infinite cyclic groups. I've read that such a group can be represented with the group of integers under addition (which makes sense). However, I was wondering if this infinite cyclic group could also be viewed through the lens of the group of real numbers under addition.

If they cannot, could someone please explain why? (My math literacy is not overly sophisticated so I would greatly appreciate simplicity if possible).

Thanks~

Answer 1

Any integer can be expressed by adding together finitely many copies of either $1$ or its inverse $-1$. $1$ (and $-1$) are said to generate the group and the group is said to be cyclic because of this. There is no generator for either the rationals or reals, so they do not form a cyclic group under addition. They do form groups under addition, but they are not cyclic.

Answer 2

Please note that the question and the title are not about the same thing.

Anyway, neither the rationals under the addition nor the reals under the same operation are cyclic. Indeed, if $x\in\mathbb R$, the group generated by $x$ is $\mathbb{Z}x$, which is different from $\mathbb R$.

Answer 3

Let’s suppose that $\mathbb{Q}$ is a cyclic group generated by some element $\frac{a}{b}$, where $a$ and $b$ are positive coprime integers. Thus by assumption we have that

$$\mathbb{Q} = \{n\left(\frac{a}{b}\right): n\in\mathbb{Z}\}.$$

However, consider some $k\in\mathbb{Z}$ such that $b<k$. Then there must exist some $n_0$ such that

$$n_0\left(\frac{a}{b}\right) = \frac{1}{k} \Longrightarrow n_0a = \frac{b}{k}.$$

But $n_0a$ is an integer, and $\frac{b}{k}$ is not! This is a contradiction, so $\frac{a}{b}$ cannot generate $\mathbb{Q}$.

Can you use a similar idea to show that, if you suppose that there exists some $x\in\mathbb{R}$ such that $\mathbb{R} = \{nx: n\in\mathbb{Z}\}$, you must arrive at a contradiction?