# 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!

-
The group $\mathbb{Z}^n/\ker \phi$ is a particular example of a quotient group (or sometimes factor group). This should be adequately explained in whatever textbook you're currently reading (you probably shouldn't even have begun to read these kinds of proofs, without understanding these notions first), otherwise have a look here: en.wikipedia.org/wiki/Quotient_group – Martin Wanvik Apr 12 '12 at 16:41
Basically, if you have an abelian group $G$ and a subgroup $H$ of it, the quotient group $G/H$ is a group whose elements are the subsets of $G$ of the form $g+H$. But you really need to study carefully a textbook in order to get a thorough understanding. – Andrea Mori Apr 12 '12 at 17:29
Ah thank you I did know about quotient groups I just have never seen it denoted like that :) – user26069 Apr 12 '12 at 22:46

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.

-