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 $\{\frac{a}{2^{n}}:a,n \in \mathbb{Z} \}$.
Best Answer
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?