$$M = \left\{\,\dfrac{m}{13^n}\biggm| m\in \mathbb{Z}, n\in\mathbb{N} \,\right\}, \quad G = M/\mathbb{Z}$$
Prove that any subgroup $H < G$, $H\neq G$ is cyclic and infinitely generated and that $G/H$ is isomorphic to $G$.
I can't find a generator for $H$; if you could give me some ideas, I would really appreciate.
What I have tried so far: factoring the group : factoring: $G=\{1^*,1/13^*,{1/13^2}^*,{1/13^3}^*,\ldots\}$ where $^*$ stands for class.
On
Hints:
(1) $$H\subsetneqq G\implies \frac1{13^n}+\Bbb Z\notin H\;,\;\;\text{for some}\;\;n\in\Bbb N$$
Let $\;n\in\Bbb N\;$ be the minimal such power, meaning $\;\frac1{13^k}+\Bbb Z\in H\;\;\forall\,k<n\;$ . Show now that
$$H=\left\langle \;\frac1{13^{n-1}}+\Bbb Z\;\right\rangle\cong C_{13^{n-1}}$$
(2) Any finite number of elements from $\;M/\Bbb Z$ involves only a finite number of powers of $\;13\;$ in the denominators...
(3) If $\;H\;$ has the form described above, define
$$\phi:G\to G\;,\;\;\phi(\alpha):=13^{n-1}\alpha\;,\;\;\alpha\in G=M/\Bbb Z$$
and now show what has to be shown and apply the FIT(=First Isomorphism Theorem)
For $g\in G$, let $f(g)=\min\{\,n\in\mathbb N_0\mid \exists m\colon \frac m{13 }\in g\,\}$. Show that $$\tag 1g\in\langle h\rangle \iff f(g)\le f(h).$$ Thus if $f$ is unbounded on $H$ then $H=G$. So for a proper subgroup $H$ we find $h\in H$ that maximizes $f$ and hence is a generator of $H$. Oberve that we may choose the generator $h$ in the form $\frac1{13^n}+\mathbb Z$. Conclude that multiplicaton with $13^n$ is an isomorphism $G\to G/H$.
Generalizing $(1)$, show that $$ g\in \langle h_1, \ldots, h_k\rangle\implies f(g)\le\max\{f(h_1),\ldots,f(h_k)\}$$ and conclude that $G$ is not finitely generated because $f$ is unbounded on $G$.