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.

Solution 1:

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.

Solution 2:

HINT $\mathbb{R}$ is uncountable

Solution 3:

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.

Solution 4:

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?