Show that $(\mathbb{R}^*, \cdot)$ is not cyclic.

Question

I'm trying to prove that the group $(\mathbb{R}^*, \cdot)$ is not cyclic (similar to [1]). My efforts until now culminated into the following sentence:

If $(\mathbb{R}^*,\cdot)$ is cyclic, then $\exists x \in \mathbb{R}^*$ such that $x \cdot x \neq x \in \mathbb{R}^*$.

The assumption written above is only not true for the neutral element on $\mathbb{R}^*$. Are there any follow-ups that I should do to improve that sentence?

[1] Show that (Q, +) and (R, +) are not cyclic groups.

Answer 1

Note that for any $n \in \Bbb N, -1$ has no $n$th root in $\Bbb R^*$ except (when $n$ is odd) $-1$ itself. Therefore, if $-1 \neq x \in \Bbb R^*, -1 \notin \langle x \rangle$. Since $-1$ also does not generate $\Bbb R^*, \Bbb R^*$ is not cyclic.

Answer 2

Hint: If $G$ is a cyclic group and $H\cong G$, then $H$ is also cyclic. Now use the fact that $(\Bbb R, +)$ is not cyclic.

Also, every group satisfies $x^2\neq x$ for each $x\neq e$. Indeed, if $x^2=x$, then $x^2=ex$, so, multiplying by $x^{-1}$ on the right, we get $x=e$.

Answer 3

Suppose $(\mathbb{R}^*, \cdot)$ is cyclic then there exist $x\in \mathbb{R}^*$ such that $\mathbb{R}^* = \langle x \rangle =\{x^n \vert \, n \in \mathbb{Z}\}$. However we see that the set $\{x^n \vert \, n \in \mathbb{Z}\}$ is countable but $\mathbb{R}^*$ is uncountable which is a cotradiction.

A similar argument shows that $\mathbb{R}^*$ can not be generated by a finite number of elements , that is for any reals $x_i$ ($i =1,2,\ldots ,n$) we must have $\mathbb{R}^* \ne \langle x_1, x_2 , \ldots , x_n \rangle$ for all positive integer $n$.

Answer 4

Suppose $(\mathbb{R}^*,\cdot)$ is cylclic. Now $\langle2,3\rangle$ is a subgroup of $(\mathbb{R}^*,\cdot)$, but it is cyclic since it is a subgroup of a cyclic group and therefore $\langle2,3\rangle=\langle a\rangle$ for some $a\in\mathbb{R}^*$. Now $a^n=2$ and $a^m=3$ for some $n,m\in\mathbb{Z}-\{0\}$, which imples $$a^{nm}=2^n=3^m$$ But $2$ and $3$ are prime and it is a contradiction.