# Defining the rank for a finitely generated abelian group

So first some definitions. Let $G$ be an abelian group, a basis for $G$ is a linearly independant subset that generates $G$. We say that $G$ is finitely generated if a basis for $G$ is finite.

Now there's an important theorem

Theorem: An abelian group is free abelian if and only if it has a basis.

So if $G$ is finitely generated then $G$ is free abelian. Now if $H$ is a free abelian group with a finite basis we define the rank of $H$ to be the number of elements in any finite basis for $H$.

So from all the above definitions, it seems that the notion of rank is defined for any abelian group that is finitely generated. However in my textbook Introduction to Topological Manifolds by John Lee the following is stated:

"we need to extend the notion of rank to finitely generated abelian groups that are not necessarily free abelian"

But by the theorem above, I don't see how any finitely generated abelian group cannot be free abelian. An abelian group $K$ that is finitely generated has a finite basis and so by the above theorem is free abelian.

What does the author mean in this case?

My apologies if by mistake I have misquoted/misunderstood any content from the above mentioned book (which I consider to be a really great book!)

EDIT: I incorrectly assumed that $G$ is finitely generated implies that a basis for $G$ is finite.

• The definition appears to be incomplete. For the group, $G$, of order $2$, if $g$ is the non-trivial element, isn't $\{g\}$ a "linearly independent subset that generates $G$"? Similarly, the claim "If $G$ is finitely generated then it is free abelian" is obviously false without more information. – lulu Aug 17 '18 at 16:36

We say that G is finitely generated if a basis for G is finite.

This isn't the usual definition of finitely generated. Usually, $G$ is said to be finitely generated if there exist $e_1, ... , e_n \in G$ such that every element of $G$ can be expressed as

$$g = c_1e_1 + \cdots + c_ne_n$$

for some $c_i \in \mathbb Z$. This doesn't mean that $\{e_1, ... , e_n\}$ or any subset of it has to be a basis. There exist finitely generately abelian groups which do not have any basis.

Example: $G = \mathbb Z/3\mathbb Z \times \mathbb Z/3\mathbb Z$ is finitely generated, by $e_1 = (1,0)$ and $e_2 = (0,1)$. But no subset of $\{e_1, e_2\}$ is a basis for $G$.

• I assumed that $G$ is finitely generated implies that a basis for $G$ is finite, my apologies, thanks for this answer! – Perturbative Aug 17 '18 at 16:43
• And for example $\{e_1, e_2\}$ is not a basis for $G$ because it isn't linearly independant since $$3(1, 0) + 3(0, 1) = (0, 0) = 0_{\mathbb{Z}/3\mathbb{Z} \times\mathbb{Z}/3\mathbb{Z} }$$ – Perturbative Aug 17 '18 at 16:57
• Right $\space$ – D_S Aug 17 '18 at 17:02