Betti numbers and Fundamental theorem of finitely generated abelian groups

Question

My textbook (author : fraleigh) says that Fundamental theorem of finitely generated abelian groups

Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$\Bbb{Z_{(p_1)^{r_1}}}\times \Bbb{Z_{(p_2)^{r_2}}}\times\dots\times \Bbb{Z_{(p_n)^{r_n}}}\times\underbrace{\Bbb{Z}\times\Bbb{Z}\dots\times\Bbb{Z}}_{\text{r times, r : betti number}}$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^{r_i}$ are unique.

I don't understand the theorem. So I was trying to show examples. So I can apply the theorem to several problems. But I don't know what Betti number say. Can you tell me some examples to understand Betti numbers? Actually when I see the next statements, I cannot start.

Any two finitely generated abelian groups with the same Betti number are isomorphic.

It is false.

Every finite abelian group has a Betti number of 0.

It is true.

Answer 1

The appearance of the term "Betti number" in your theorem should be seen as a definition. The Betti number of a finitely generated abelian group $G$ is the number of $\mathbb{Z}$ factors appearing in the decomposition claimed by your theorem, or differently stated, the rank of $G$ as a $\mathbb{Z}$-module.

You can also understand it as a sort of analogue to the dimension of a vector space. Indeed, tensoring $G$ with $\mathbb{Q}$ will convert it into a $\mathbb{Q}$-vector space whose dimension is exactly the rank/Betti number of $G$.

That is, for example, if $G=\mathbb{Z}^n$, then $n$ is the Betti number.

Another example is $G=\mathbb{Z}_{(p)}$. There is no $\mathbb{Z}$ factor, so the Betti number is $0$.

Now let us look at your two statements:

The first one is false: Take $G=\mathbb{Z}$ and $H=\mathbb{Z}\times \mathbb{Z}_{(2)}$. They both have Betti number $1$, but they are clearly not isomorphic.

The second one is true: if the Betti number was positive, then the group would have a subgroup isomorphic to $\mathbb{Z}$, in particular it would have infinitely many elements.