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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?2012-09-29

7 Answers 7

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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 $\mathbb R$.2014-08-15
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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...2012-09-29
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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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    If you wonder, I already upvoted ;)2012-09-30
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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?2013-12-25
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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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$\mathbb{R}^*$ 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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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 $G\neq \mathbb R^*$.