What does factorization of a group mean? I've decided to come here after looking for days in books and online about what factorization for groups means in algebra. More specifically, the following group (with +)    is given: 
M={ m/13^n | m in Z, n in N}, Z is a subgroup of M, M is a subgroup of Q. And G is given as M/Z (which supposedly translates to M factorized to Z).
And the question is, if H is a subgroup of G and H != G (not equal) ==> H is cyclic.
So, I can't even begin to think how to solve this problem if I don't even know what M/Z means. I've looked this up in a lot of books, and it is used quite often but they never explain what it means.
Also, if you could give me some hints, this problem has 2 more subpoints:
i) G is not finitely generated
ii) H is a subgroup of G ==> G/H ~ G
 A: The notation $M/Z$ means that you are "factoring out" $M$ by the equivalence relation $a \sim b$ if $a-b \in Z$.   i.e, $M/Z$ is the set of equivalence classes of this equivalence relation under the operation $[a]+[b]=[a+b]$ (where $[a]$ denotes the equivalence class of $a$).
So for instance. Let $G= \mathbb{Z}/ 4 \mathbb{Z}$ and let $H= 2 \mathbb{Z}/4 \mathbb{Z}$. Then $G/H$ has two elements given by the equivalvence classes $[1], [0]$ and we see that if $\mathbb{Z}/ 4 \mathbb{Z}= \{ 0, 1, 2, 3\}$ then $[1]=\{1,3\}, [0]=\{2,4\}$ and $[1]+[1]=[2]=[0]$. Thus, $G/H \equiv \mathbb{Z}/2 \mathbb{Z}$ in this case. 
A: $G/H$ is called a quotient group as others have said. Here's some intuitive motivation; I'll leave the rigorous details to textbooks.
A good motivating example is $\mathbb{Z}/5\mathbb{Z}$, the integers mod $5$ under addition. Think of it as the integers with the extra relation $5=0$, so there are just the numbers $\{0, 1, 2, 3, 4\}$ with eg. $3+4=2$. Setting $5=0$ forced us to set $5n=0$ for all $n \in \mathbb{Z}$. Indeed, we're setting precisely the elements of $\langle 5 \rangle = 5 \mathbb{Z} \subset \mathbb{Z}$ equal to zero and then fixing up the addition table to be consistent with this requirement.
More generally, given a group $G$ and an element $h \in G$, setting $h=1$ (I'll use multiplicative notation now) forces us to set $h^n = 1$ for $n \in \mathbb{Z}$, so all of $\langle h \rangle$ is $1$. Indeed, for any $g \in G$, we also need $ghg^{-1} = gg^{-1} = 1$, so we're typically setting "more" than $\langle h \rangle$ to $1$. If $G$ is abelian, as in the previous example, $ghg^{-1} = gg^{-1}h = h$, so we can get away with just setting all of $\langle h \rangle$ to $1$.
In general, setting some elements of $G$ to $1$ has the effect of setting an entire "normal subgroup" $H \subset G$ to $1$; this is a subgroup such that $ghg^{-1} \in H$ for all $g \in G$. Any normal subgroup can be set to $1$ without setting anything not in that subgroup to $1$. So, when considering which groups arise as quotients of some initial group, we just look for normal subgroups. Note that $\{1\}$ and $G$ are trivially normal subgroups of $G$, corresponding to the fact that we can always set only $1$ to $1$ (i.e. do nothing) or we can set everything to $1$ (resulting in the one-element group nobody wants).
A very important class of groups are those for which the only way to set some elements to $1$ is trivially, i.e. whose normal subgroups are just the one-element group and the entire group. Such groups are called "simple". The finite simple groups have been classified (most likely; it's extremely long).
