Prove a finitely generated abelian group in which every element is of finite order is finite.

Question

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.

Answer 1

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