# Prove that $\mathbb{R^*}$, the set of all real numbers except $0$, is not a cyclic group

Prove that $\mathbb{R^*}$ is not a cyclic group. (Here $\mathbb{R^*}$ means all the elements of $\mathbb{R}$ except $0$.)

I know from the definition of a cyclic group that a group is cyclic if it is generated by a single element. I was thinking of doing a proof by contradiction but then that ended up nowhere.

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Hint: can $-1$ be any power of another real number? –  Thomas Andrews Sep 29 '12 at 21:36

Suppose $\mathbb{R}^*$ is cyclic. Let $a$ be its generator. Since $-1 \in \mathbb{R}^*$, there exists a nonzero integer $n$ such that $-1 = a^n$. Then $a^{2n} = 1$. Hence the order of $a$ is finite. This is a contradiction.

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$a^{2n}=1$ does not imply $2n=0$. for example $a=1,2n=10$. though in general this proof is correct since you deduce $a=-1$ from $a^n=-1$ –  Belgi Sep 29 '12 at 22:23
@Belgi If $a^{2n} = 1$ for some $n \neq 0$, $\mathbb{R}^*$ must be finite. –  Makoto Kato Sep 29 '12 at 22:28
After the edit it is correct, I objected to the $2n=0$ part that seemed to be concluded only since $a^{2n}=1$ –  Belgi Sep 29 '12 at 23:30
I agree with the edited version. I just wanted to say that without saying that otherwise the group would of been finite you can't deduce straight out that $2n=0$. thank you –  Belgi Sep 30 '12 at 8:13
Yes, I just meant this argument should be inserted in the proof –  Belgi Sep 30 '12 at 9:35

HINT $\mathbb{R}$ is uncountable

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That's the extreme answer, but there are more algebraic answers that also let you show that for any sub-field of the reals the non-zeros are not cyclic... –  Thomas Andrews Sep 29 '12 at 21:43

clark's answer is surely a great and simple one. Thomas Andrews' hint is another great one. Here's a more complicated answer that shows more, that there are entire intervals of numbers that would not be generated.

Let $x$ be a generator of the cyclic group $\mathbb{R}^*$. If $|x| = 1$, then all powers of $x$ satisfy $|x^n| = 1$. So, $|x| < 1$ or $|x| > 1$.

If $|x| < 1$, then $|x^{-1}| > 1$ and $x^{-1}$ is also a generator. So, assume $|x| > 1$.

If $|x| > 1$, then $|x| = 1 + \epsilon$ for some $\epsilon > 0$. Any positive power of $x$ will satisfy $|x|^n = (1 + \epsilon)^n > 1 + \epsilon$. Any negative power of $x$ will satisfy $|x|^{-n} = (1 + \epsilon)^{-n} < (1+ \epsilon)^{-1}$.

So, there are entire intervals that are never achieved.

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I upvoted. Can you please flesh out where the last sentence came from? –  François Muer Dec 25 '13 at 7:22

Say $g$ is the generator. It must be negative. Either $g<-1, g=-1,$ or $g>-1$. $g=-1$ clearly does not work. Let $h = \frac12(-1 + g)$. $h$ lies strictly between $g$ and $-1$. How is $h$ generated?

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Let $g\in\mathbb R^*$ and $G=\langle g \rangle=\{ g^n : n \in \mathbb Z\}$.

If $|g|=1$ then $G \subseteq \{ \pm 1 \} \neq \mathbb R^*$.

Otherwise, we may assume that $|g|>1$.

If $g>1$ then $x=(g+1)/2 \notin G$ because $1 < x < g$.

If $g<-1$ then $x=(g-1)/2 \notin G$ because $g < x < -1$.

In both cases, we have $G\neq \mathbb R^*$.

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$\mathbb{R}^*$ has infinite order. If it were cyclic, it would have to contain an element that does not have a cube root.

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Another way to see it: Assume $\Bbb{R}$ is cyclic and $a$ is its generator. Then $1=na$ for some $n\in\Bbb{Z}$ and $\sqrt{2}=ma$ for some $m\in\Bbb{Z}$. These imply that $a=1/n=\sqrt{2}/m$ and hence $\sqrt{2}=n/m$, contradiction.

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The question is about the multiplicative group $\mathbb R^*=\mathbb R\setminus\{0\}$, not the additive group $\mathbb R$. –  Christoph Aug 15 '14 at 14:04