Factor/Quotient Group $M/\mathbb{Z}$ I've posted another question here a few days ago asking what a factor/quotient group is because I couldn't wrap my mind around it. Although I have an idea what it means, I still don't fully understand it and why we use it. So can you please tell me what the following factor group means explicitly:
$M=\left\{ m\times13^{-n}\mid m\in\mathbb{Z}\wedge n\in\mathbb{N}\right\} $
with $\mathbb{Z}\leq M\leq\mathbb{Q}$
So how does $M/\mathbb{Z}$ look, what does factoring to $\mathbb{Z}$ mean?
I need to prove that a subgroup $H\leq M/\mathbb{Z}$ is cyclic.
(don't start with $\mathbb{Z}/4\mathbb{Z}$ or examples like that, i already got that)
 A: As a heuristic, the factor group $G/H$ is what you get when you want to consider elements of $G$, but don't really care about difference by elements of $H$. 
As an example, in calculus, when studying the behavior of a function at the origin, it's often useful to consider it up to first order, i.e. write $f(x) = a_0 + a_1x + x^2( \ldots)$.  So here $G$ could be power series in $x$ under addition, and $H$ could be power series with $a_0 = a_1 = 0$; in the quotient $G/H$ we just take functions up to first order, i.e. from $f$ we remember only $a_0 + a_1x$. 
As another example, in the Dihedral group $D_{2n}$, you could be only interested in whether your symmetry preserves or reverses orientation. In this case, since rotations preserve orientation, you would want $D_{2n}/\mathbb{Z}_n \cong \mathbb{Z}_2$.
As to your examples, $M$ is plainly just rational numbers with only powers of 13 in the denominator. As Karl Kronenfeld helpfully says in the comments, when you don't care about difference by $\mathbb{Z}$, all you are left with is the fractional parts (mod 1).
Let's see any (proper) subgroup $H$ is cyclic. Suppose the fractional parts of elements of $H$ get arbitrarily close to 0, it follows that the powers of 13 in the denominators of $H$ grow arbitrarily big, i.e. we have $\frac{a}{13^k} \in H$ with $(a, 13) = 1$, and $k$ arbitrarily large.  Since $(a,13^k) = 1$, we can write $an + 13^km = 1$, with $n,m \in \mathbb{Z}$, so we also have $\frac{na}{13^k} + \frac{m13^k}{13^k} = \frac{1}{13^k}$ in $H$, i.e. $H =G$.
So any proper subgroup $H$ has bounded denominators, say all bounded by $13^k$. It follows that $H$ is a subgroup of the one generated by $\frac{1}{13^k}$, which is isomorphic to $\mathbb{Z}/13^k \mathbb{Z}$. Since a subgroup of a cyclic group is cyclic, we're done :).      
