Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic

Question

I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"

Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.

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Here is my attempt for the first.

Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.

$\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.

To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.

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However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible.

Answer 1

A nicer proof, perhaps, is to note that if $\Bbb Q^\times$ were cyclic, being infinite, must isomorphic to $\Bbb Z$. But $\Bbb Z$ has no element of order $2$, whereas $(-1)^2=1$ in $\Bbb Q^\times$. Note that this proves then that $\Bbb R^\times$ cannot be cyclic either.

ADD To be more precise, $$\Bbb Q^\times \simeq \Bbb Z/2\Bbb Z \oplus \bigoplus_{i\geqslant 1}\Bbb Z$$ by using the prime factorization.

Answer 2

Hint: If ${\mathbb R}$ is cyclic then it's countable. Your argument for ${\mathbb Q}$ looks good. Also a subgroup of a cyclic group is cyclic so you could go that route for ${\mathbb R}$ too, based on your proof for ${\mathbb Q}$.

Answer 3

Simple proof:

If $|x|>1$, then $|x^n| > 1$ for all $n \in \mathbb{N}$.

If $|x|<1$, then $|x^n|< 1$ for all $n \in \mathbb{N}$.

So take a purported generator $g$. Well, $|g| \neq 1$, so either $|g|<1$, or $|g|>1$. In the former, we won't generate any numbers with magnitude larger than $1$. In the latter, we will not generate any numbers with magnitude less than $1$. Thus, neither group can be cyclic.

Answer 4

Here's another way to argue these points:

Suppose $\Bbb Q^\ast$ were cyclic; then $\Bbb Q^\ast = \{ a^n, n \in \Bbb Z \}$ for some $a \in \Bbb Q^\ast$. If $\vert a \vert = 1$, then $a = \pm 1$, and $\langle a \rangle = \{ 1 \} \;\text{or} \; \{ \pm 1 \}$, so $\vert a \vert \ne 1$. If $\vert a \vert > 1$, the sequence $\vert a^n \vert = \vert a \vert^n$ is monotonically increasing; thus there can be no $m \in \Bbb Z$ with $\vert a^n \vert < \vert a^m \vert < \vert a^{n + 1} \vert$; thus we cannot have $\Bbb Q^\ast = \langle a \rangle$, since there is always a rational between $\vert a^n \vert$ and $\vert a^{n + 1} \vert$. The case $\vert a \vert < 1$ may be similarly argued, but with $\vert a^n \vert$ decreasing. So $\Bbb Q^\ast$ is not cyclic.

Essentially the same argument may be applied to $\Bbb R$. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 5

Your proof is good and as noted by user2566092 the fact that $\Bbb R^\times$ is uncountable implies immediately that it cannot be cyclic.

Here is a different approach to $\Bbb Q^\times$:

If $\Bbb Q^\times$ were cyclic it would certainly be infinite cyclic. But we know that all non-trivial subgroups of a cyclic infinite subgroup are infinite, and $\Bbb Q^\times$ has a finite non-trivial subgroup, namely $\{1,-1\}$.

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Beat by PedroTamaroff by a few seconds :)

Answer 6

Simpler form of your proof: If $(\mathbb{Q}^\times, \cdot)$ is cyclic, it has a generator $\frac{p}{q}$ in lowest terms. Now $\frac{2 p}{q} \in \mathbb{Q}^\times$, so it there must be $n \in \mathbb{Z}$ such that: \begin{align} \frac{2 p}{q} &= \left( \frac{p}{q} \right)^n \\ 2 q^{n - 1} &= p^{n - 1} \end{align} (we consider $n > 0$ here, switch around $p$, $q$ so both sides are integers if $n < 0$). So $p$ is even, and $q$ must be odd. The only possibility is to have $n = 2$, and so $q = 1$, and thus $p = 2$, the generator must be $\frac{1}{2}$. But there is no way to get $\frac{1}{3}$ as a power of $\frac{1}{2}$.