Exercise 2.4.15 (Proper subgroup of $\mathbb{Q}$ which is not cyclic)

Proof. Consider the set $𝔻$ of decimal numbers:

Then $𝔻\subseteq \mathbb{Q}$ and $0=0∕10^0\in 𝔻$, so $𝔻\ne \varnothing$.

Moreover, if $x=a∕10^k\in 𝔻$ and $y=b∕10^l\in 𝔻$, then

so $x-y\in 𝔻$.

This shows that $𝔻$ is a subgroup of $\mathbb{Q}$.

Assume, for the sake of contradiction, that $𝔻$ is cyclic. Then there exists some fixed $x_0=a_0∕10^{k_0}\in 𝔻$ such that $𝔻=⟨x_0⟩$. Since $1∕10^{k_0+1}\in 𝔻$, there is some $n\in \mathbb{Z}$ such that

thus $1=10n{a}_0$, where $n{a}_0\in \mathbb{Z}$, so $10\mid 1$. Since $10\nmid 1$, this contradiction proves that $𝔻$ is not cyclic.

Note that $𝔻\ne \mathbb{Q}$, otherwise $1∕3\in 𝔻$. Then there exist $q,a\in \mathbb{Z}$ and $k\in \mathbb{N}$ such that

Therefore $10^k=3\mathit{qa}$, so $3\mid 10^k$. Since $g.c.d.(3,10)=1$, then $g.c.d.(3,10^k)=1$. This implies $3\mid 1$, which is false. Hence

In conclusion, $𝔻$ is a proper subgroup of $\mathbb{Q}$ which is not cyclic.

(Therefore $𝔻$ is not finitely generated by Exercise 14.) □