0
$\begingroup$

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~

$\endgroup$

closed as unclear what you're asking by Matthew Towers, user99914, Saad, The Phenotype, Xander Henderson Mar 23 '18 at 2:44

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ You ask about rationals in the title and reals in the text. The answer here is the same, but the number sets are not. $\endgroup$ – Ross Millikan Mar 22 '18 at 16:39
  • $\begingroup$ If the reals or the rationals are cyclic, what's their generating element? $\endgroup$ – user4894 Mar 22 '18 at 16:39
  • $\begingroup$ To the OP: do yourself a favor and edit your question to ask whether the infinite cyclic group is "isomorphic to the group of rational numbers under addition" rather than "can be viewed through the lens of...". Please remove the real numbers from the discussion. They are uncountable so cannot be isomorphic to a countable group. $\endgroup$ – fredgoodman Mar 23 '18 at 16:42
  • $\begingroup$ I voted to re-open, since the OP clearly intended to ask whether the reals (or rationals) with addition are infinite cyclic groups. With just a little generosity, one can interpret this question well enough to provide an instructive answer. $\endgroup$ – fredgoodman Mar 23 '18 at 19:15
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ so, basically, you need some sort of 'base unit' that builds up all of the other elements of the set (which for the integers is 1 and -1). But for real numbers or rational numbers (sorry about the title mishap...meant to ask about both), for any 'suggested' base unit, I could find a smaller one. Therefore, there is no specific base unit. Am I close? $\endgroup$ – S.Cramer Mar 22 '18 at 16:55
  • $\begingroup$ You are right to be thinking about the base unit. That is the generator. For both reals and rationals there are two problems. Being able to find a smaller one is one, but you also miss a lot of them. In the rationals, if your generator is $r$ you can't express $\frac 32r$, for example. In the reals, there is that problem and you can't express $\sqrt 2 r$ $\endgroup$ – Ross Millikan Mar 22 '18 at 17:01
1
$\begingroup$

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 $\mathbb R$.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.