# Finitely generated and torsion free abelian group is isomorphic to Z^d.

I have to prove the following:

Let $(G,+)$ be a finitely generated and torsion-free abelian group then $G\cong \mathbb{Z}^d$ where $d$ is the cardinality of the smallest possible generating set of $G$.

I've come up with a proof that shows $G\cong \mathbb{Z}^d$ but I dont use the fact that $G$ is a torsion-free group which is odd to say the least.

Is this hypothesis necessary ? Have I made a careless mistake in my proof ?

Here is the proof:

$G = \langle g_1, g_2, ..., g_d\rangle$

Let $f : G \rightarrow \mathbb{Z}^d : g = \sum_{i = 1}^d a_ig_i \mapsto (a_1,a_2, ..., a_d)$.

Then $f$ is a homophormism:

Let $g' = \sum_{i=1}^d a_i'g_i \in G \;$and $g = \sum_{i = 1}^d a_ig_i \in G$

$$f(g+g') = (a_1 + a_1', ...,a_d + a_d') = (a_1,a_2, ..., a_d) + (a_1',a_2', ..., a_d') = f(g) + f(g')$$

We have $\operatorname{Ker}(f) = \{ g \in G | \; f(g) = 0 \} = \{e \}$:

Let $g = \sum_{i = 1}^d a_ig_i \in\operatorname{Ker} (f)$ then $$f(g) = (a_1, ...,a_d) = (0, ..., 0) \iff a_i = 0 \; \forall i = 1,2 , ...,d \Rightarrow g = 0.$$ Therefore $\operatorname{Ker}(f) \subseteq \{0\} \Rightarrow\operatorname{Ker}(f) = \{0\}$ and f is an injective function.

We have that $\operatorname{Im}(f) = \mathbb{Z}^d:$

Let $z =(z_1,z_2, ...,z_d) \in \mathbb{Z}^d$ and define $g = \sum_{i = 1}^d z_i g_i$ and $f(g) = z$

Therefore $f$ is an isomorphism and $G \cong \mathbb{Z}^d$

• I think that in the proof that $f$ is a homomorphism you are implicitly assuming that there are no relations between the $g_i$. I’m not sure you can immediately conclude that $f(g+g’) = (a_1 + a_1’, \dots, a_d + a_d’)$. – Santana Afton Jan 6 '18 at 15:18
• Either that or in the proof that the kernel is trivial. Yeah, I suppose for $g_i\ne 0$, it’s true that $ag_i = 0$ if and only if $a= 0$ because $G$ is torsion-free. – Santana Afton Jan 6 '18 at 15:24
• @SantanaAfton doesn't the definition of $d$ as the cardinality of the smallest possible generating set of G remove any $g_i$'s that are related ? for example if $\{ g_1,g_2,g_3\}$ is a generating set of $G$ and suppose that d = 3. Then if we have $g_3 = g_2 + g_3$ then $\{ g_1 , g_2 \}$ is also a generating set of $G$ and $d < 3$ We could in this way eliminate all related $g_i'$ until they are all "independent" and we obtain the generating set of minimum cardinality. – Digitalis Jan 6 '18 at 15:34
• You definition $f : G \rightarrow \mathbb{Z}^d : g = \sum_{i = 1}^d a_ig_i \mapsto (a_1,a_2, ..., a_d)$ define a function on $G$ if and only if $\{g_i\}$ is a base for $G$ as $\Bbb Z$-module because you are assuming that each element $g$ of $G$ is of the form $\sum_{i = 1}^d a_ig_i$ for unique $(a_1,a_2, ..., a_d)$. – Fabio Lucchini Jan 6 '18 at 15:38
• @FabioLucchini I'm not sure I understand what you mean. I'm not familiar with what a base is in group theory but if it similar to the concept in linear algebra $\{g_i | i = 1,2,...d\}$ is a generating set of minimal cardinality(by definition of $d$) so it is something like a base. However I dont know what a $\mathbb{Z}$ module is ... – Digitalis Jan 6 '18 at 15:48

You have not proved that the function $f : G \to \mathbb{Z}^d$ is well-defined.
And if you apply your definition of $f$ to simple counterexamples, you will see that it is not well-defined in general.
Take, for example, the direct sum of two cyclic groups of order $2$ with generators $g_1,g_2$. Letting $g=g_1$ we have $$g = 1 \cdot g_1 + 0 \cdot g_2$$ and so $f(g)=(1,0)$.
But wait! We also have $$g=3 \cdot g_1 + 4 \cdot g_2$$ so $f(g)=(3,4)$.