Trouble with a proof for a finitely generated abelian group. Let $G$ be a finitely generated abelian group. Then there exists an integer $n \geq 0$ and subgroup $H$ of $\mathbb{Z}^n$ such that $G \cong \frac{\mathbb{Z}^n}{H}$. 
The proof constructs surjective homomorphism $\phi : \mathbb{Z}^n \to G$ by mapping $(x_1,\dots,x_n) \in \mathbb{Z}^n$ to $x_1g_1 + \dots + x_ng_n \in G$ for generators $g_1,\dots,g_n$ of $G$. Then says by the First Isomorphism Theorem, we have $G \cong \frac{\mathbb{Z}^n}{\ker \phi}$
Firstly I don't understand this last 'jump', and also what does it mean to divide groups like that? Any help would be appreciated!
 A: Let $G$ be an abelian group and $H$ a subgroup of $G$. Consider the relation on $G$ which is given by $a\sim b$ $:\Leftrightarrow$ $\exists h\in H: a+h = b$. This is an equivalence relation:


*

*If $a+h=b$ for some $h$, then $b+(-h)=a$ and $-h\in H$, so it is symmetric.

*If $a+h=b$ and $b+g=c$, then $a+(h+g)=c$ - and for $g,h\in H$ we have $g+h\in H$. Thus, it is transitive.

*Clearly, $a+0=a$ and $0\in H$.


Therefore, we denote by $[a]=\{ g\in G \,\vert\, g\sim a\}$ the equivalence class of some group element $a\in G$. We then denote the set of all these equivalence classes by $G/H=\{ [g] \,\vert\, g\in G\}$. Define a group operation on $G/H$ by the law $[a] + [b] := [a+b]$ and check that this is well-defined and an abelian group operation:


*

*Let $[a]=[a']$ and $[b]=[b']$ and pick $h,f\in H$ with $a+h=a'$ and $b+f=b'$. Then, $(a+b)+(h+f)=a'+b'$, so $[a]+[b]=[a']+[b']$ as required.

*The validity of the group laws and commutativity follow automatically.


Now, for the infamous
First Isomorphism Theorem. Let $G$ and $H$ be as above and consider a morphism of abelian groups $\phi:G \to M$. Assume that $H\subseteq\ker(\phi)$, then there is a morphism $\psi: G/H\to M$ with $\psi([a])=\phi(a)$ for all $a\in G$. Furthermore, if $H=\ker(\phi)$, then $\psi$ is injective.
Proof. Define $\psi$ as stated and check that it is well-defined: If $[a]=[a']$, we pick $h\in H$ with $a+h=a'$, then $\phi(h)=0$ by assumption and thus, $$ \psi([a])=\phi(a)=\phi(a)+\phi(h)=\phi(a+h)=\phi(a')=\psi([a']). $$ 
Now, if $H=\ker(\phi)$, then $\psi([a])=0$ if and only if $a\in H$, which is the case if and only if $[a]=[0]$, so $\psi$ is injective. qed
If we now also add the assumption that $\phi$ is surjective, as in your example, then we get a map $\psi: G/H \to M$ which is both injective and surjective, hence an isomorphism.
