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Let $A$ and $B$ are free $\mathbb{Z}$-module and has the same rank $n$. If rank $A/B＝0$, then $|A/B|$ is finite ?

From rank $A/B＝0$, I could deduce that orders of every element $A/B$ is finite. But from this, I think

“$|A/B|$ is finite” does not follow immediately. How can I follow the logic between them ?

Thank you for your kind help.

Farshad Hasani
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    If $n$ is finite, then $A$ is finitely generated, so $A/B$ is a finitely generated abelian group in which every element has finite order, hence is finite. If $n$ is infinite, then it does not follow: take the direct sum of countably many copies of $\mathbb{Z}$, for $A$, and take $B$ to be the direct sum of the submodules $2\mathbb{Z}$. Then the quotient has zero rank, but is the direct sum of countably many copies of $\mathbb{Z}/2\mathbb{Z}$, which is infinite. – Arturo Magidin Oct 25 '19 at 16:05
  • Thank you for your comment! I understand that the proposition does not hold in infinite case.But every element has finite order, then, why the order of group is finite? I cannot follow the logic here. I'd be appreciated if you write answer in answer page. –  Oct 25 '19 at 16:12
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    I explained why: if $n$ is finite, then $A$ is finitely generated, so the quotient $A/B$ is finitely generated. A finitely generated abelian group in which every element is of finite order is finite. If $a_1,\ldots,a_n$ are generators, and they are of order $k_1,\ldots,k_n$, then since every element can be expressed in the form $a_1^{\alpha_1}\cdots a_n^{\alpha_n}$ with $0\leq \alpha_i\lt k_i$, then the group has at most $\prod k_i$ elements. That's finite. – Arturo Magidin Oct 25 '19 at 16:18
  • That is, I want to know the detail of ' Hence'. –  Oct 25 '19 at 16:18
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    So you are correct that "every element has finite order" is not enough, by itself to conclude the group is finite. But here you also know finite generation and abelianness. – Arturo Magidin Oct 25 '19 at 16:18
  • You are correct. Thank you very much! –  Oct 25 '19 at 16:23
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    @ArturoMagidin It may be useful if you collect your comments into an answer. I've done a little bit of searching, and - to my surprise - I didn't find a good duplicate target. (My search fu isn't great, so if you look you may easily find one, but if not it's for sure better readable as an actual answer.) – Daniel Fischer Oct 25 '19 at 19:49
  • @ArturoMagidin Great, thanks. May the grading be not too tedious. – Daniel Fischer Oct 25 '19 at 20:32

1 Answers1

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Let $n\in \Bbb N$ and let $\{g_1,...,g_n\}$ be a set of generators for the Abelian group $G$ with identity $1.$ For $1\le i\le n$ let $o_i\in \Bbb N$ with $(g_i)^{o_i}=1.$

For any $x\in G$ there exist integers $x_1,...,x_n$ such that $x=\prod_{i=1}^n(g_i)^{x_i}.$

For each $x_i$ there exists $x'_i\in \Bbb N$ with $x_i\le o_i$ such that $x'_i\equiv x_i \pmod {o_i}.$

So $x=\prod_{i=1}^n(g_i)^{x'_i}.$

The number of such products is at most $P=\prod_{i=1}^no_i.$

So $|G|\le P\in \Bbb N.$

DanielWainfleet
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