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Non cyclic subgroup of nonzero rationals under multiplication:

My motivation to question is curiosity. Moreover,this would be another proof that the set of non zero rationals under multiplication is not cyclic group.

I see that set of dyadic rationals is subgroup of rationals under addition, so, I tried to show non zero dyadic rationals is non cyclic subgroup of non zero rationals under multiplication, but turned out, it's not even subgroup.

The set of dyadic rationals is $\{\frac{a}{2^{n}}:a,n \in \mathbb{Z} \}$.

Mahmoud albahar
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3 Answers3

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The positive rationals form a subgroup, which is actually the free abelian group on a countable basis (because of unique factorization): any positive rational number can be written in a unique way as $$ \prod_{k\ge0} p_k^{e_k} $$ where $p_0=2,p_1=3,p_2=5,\dotsc$ is the sequence of distinct primes and the exponents $e_k$ are integers, all zero except for a finite number of them (so the product is actually finite).

You're not required to mention this, but it should give an idea: take the subgroup $H$ generated by $2$ and $3$. The elements of $H$ can be written in a unique way as $2^a3^b$ where $a,b$ are integers. Is this subgroup cyclic?

egreg
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Consider $G=(\Bbb{Q}\setminus \{0\} , \cdot) $ and $\textrm{sgn}:G\to C_2=\{\pm 1\}$ defined by $$\textrm{sgn}(x)=\begin{cases}1&x>0\\-1&x<0\end{cases}$$

Claim:

  1. $\textrm{sgn}$ is an onto homomorphism.

  2. $\ker(\textrm{sgn}) =\Bbb{Q}^{+}$

  3. $H=(\Bbb{Q}^{+}, \cdot)$ is not cyclic ( even not finitely generated).

Sourav Ghosh
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Note that any finite subgroup of the multiplicative group of non-zero rationals (or for that matter, non-zero elements of any field) is cyclic.

Thus, your subgroup needs to be infinite.

How about taking the subgroup generated by $\{1/p : p\text{ is an odd prime }\}$ ?

This is a proper subgroup of $\mathbb{Q}^{\times}$ as it doesn't contain $1/2$. Can it be cyclic?

Category_Theorist
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