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I'm studying the Fundamental Theorem of finitely generated Abelian group, and it says that the number of factors equal to $\mathbb Z$ (textbook says it is the Betti number of the group) is unique up to isomorphism.

So what is "the number of factors"?

I tried to find through Wikipedia, and it says that it is the number of generators. So if I'm right, the Betti number of Z_6 is 2 (since 1 and 5 are the generators). Then, what is Betti number of Z_360? Should I try all the cases that are relatively prime to 360? Is there any way to get it easier? I really want to fully understand the definition and applications. Thanks for your help :)

user26857
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Math-Nerd
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  • One additional comment; I think the "number of generators" comment on Wikipedia is slightly wrong, but at the very least it means the minimal number of generators in a generating set, not the number of elements which are generators. So the "answer" for $\mathbb{Z}_6$ would be $1$, because it is generated by $1$ (or by $5$ - but one element is sufficient). Something like $\mathbb{Z}^2$ isn't generated by a single element; but it can be generated by two elements (e.g. $(1,0)$ and $(0,1)$), so the Betti number is $2$. – mdp Apr 04 '14 at 13:01
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    Ah! Now I understand! Thank you very much! – Math-Nerd Apr 04 '14 at 13:14
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    Wait! but this textbook says every finite abelian group has a Betti number 0...? – Math-Nerd Apr 04 '14 at 13:37
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    Yes - this is why the number of generators comment on Wikipedia is wrong! The finite abelian group $\mathbb{Z}_5$ is generated by one object, but has Betti number zero (its free part is $0$). I think the confusion on Wikipedia is because they are talking about groups arising from topological spaces, and if you put enough restrictions on the space, you will only see groups isomorphic to $\mathbb{Z}^n$, so then the Betti number really is the number of generators - but this is not the case even for general topological spaces. – mdp Apr 04 '14 at 13:44
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    Said another way - in the textbook's language, the Betti number is not the number of generators, but the number of factors isomorphic to $\mathbb{Z}$ (and I agree that this is the better definition). As $\mathbb{Z}$ is not finite, no finite abelian group can have a factor isomorphic to $\mathbb{Z}$. – mdp Apr 04 '14 at 13:46
  • Eureka! Thank you again! – Math-Nerd Apr 04 '14 at 13:50
  • @user26857 The $n$th Betti number of a space $X$ is the Betti number of $H_n(X)$. – Najib Idrissi Jan 13 '15 at 09:32
  • Hello, I dont understand yet. Betti numbers is like number of generators?? For example, how can I find the betti number of $Z x Z_6 x Z_8 x Z x Z_{10} x Z$??? – Salvattore May 09 '16 at 05:01

1 Answers1

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Each finitely generated abelian group $G$ is isomorphic to $$\mathbb{Z}^n\oplus\bigoplus_{m\geq2}\mathbb{Z}_m^{k_m}$$ (where all but finitely many of the $k_m$ are $0$) by the fundamental theorem - which says something much more precise, but this is all we'll need. The Betti number is then $n$; the number of summands (or factors) that are isomorphic to $\mathbb{Z}$. Sometimes the $\mathbb{Z}^n$ is called the free part (and the other summands the torsion part), so you're looking for the number of generators of the free part.

Another more technical way to say this is that it's the dimension over $\mathbb{Q}$ of $G\otimes_{\mathbb{Z}}\mathbb{Q}$; because $\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}$ and $\mathbb{Z}_m\otimes_{\mathbb{Z}}\mathbb{Q}\cong\{0\}$, this agrees with the above calculation.

mdp
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    So the Betti number is the same as the rank? To me, this definition suggests that is the rank of the $k$-th homology group of the group. – k.stm Apr 04 '14 at 12:53
  • That page is about the Betti number for topological spaces; the $k$-th Betti number of the space is the rank of its $k$-th homology group, which deserves to be called the Betti number of the group! To calculate the homology group of a group, you need to put a topology on it; for finite abelian groups, the most obvious one is the discrete one, in which case the $0$-th Betti number is the number of elements, and all the others are $0$ - this probably isn't what you wanted! – mdp Apr 04 '14 at 12:55
  • Well, there’s always group (co-)hohomology. – k.stm Apr 04 '14 at 12:56
  • True - but the fact the OP says "Betti number" rather than "Betti numbers" suggests that this is not the point. I don't know if one of the cohomological Betti numbers will also agree with the rank. – mdp Apr 04 '14 at 12:59
  • BTW, then if two finitely generated Abelian groups have same Betti number, are they isomorphic? – Math-Nerd Apr 04 '14 at 13:02
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    @Math-Nerd No, $ℤ$ and $ℤ × ℤ/2ℤ$ are not isomorphic. – k.stm Apr 04 '14 at 13:02
  • For calculate Betti number... need I compute the number of isomorphic groups? – Salvattore May 09 '16 at 20:24