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

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 hypotheses of $p$ and $q$ being coprime.

However I don't know wether 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 toward a proof, without telling me the whole proof, if possible.

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Looks good to me and Subgroups of cyclic groups are cyclic. Since your proof and $\Bbb{Q} \leqslant \Bbb{R}$, $\Bbb{R}$ cannot be cyclic. – Enjoys Math Apr 10 '14 at 18:46
A detail: You may rewrite your proof as below to avoid the use of logarithm. The idea is the same, but the use of the Fundamental Theorem of Arithmetic makes the proof easier to adapt to other algebraic contexts: from $\frac{2p}{q}=(\frac{p}{q})^n$ and the fact that $p$ and $q$ are non zero, we have $2q^{n-1}=p^{n-1}$. So $2$ is a factor of $p^{n-1}$. But $2$ being prime, this means that $2$ divides $p$ and $2$ appears in the factorization of $p^{n-1}$ with a power at least $n-1$. But this implies that $2$ divides $q$, hence a contradiction. – Taladris Apr 30 '14 at 1:23
up vote 7 down vote accepted

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.

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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 ${\mathbb Q}$.

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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.

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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 $\{1,-1\}$.

Beat by PedroTamaroff by a few seconds :)

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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 $\frac{1}{2}$.

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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!!!

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Robert, I think it has been stated that answers are noT to be signed. – Pedro Tamaroff Apr 10 '14 at 19:23
@PedroTamaroff: where? – Robert Lewis Apr 10 '14 at 19:25
I think it was the blog. – Pedro Tamaroff Apr 10 '14 at 19:26

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