Proof Finitely Generated Abelian Group with no element of finite period (apart unit element), is free.

Question

I have been trying to work on a proof on why finitely generated abelian groups with no finite period element besides the identity are free (no element finite period), and I am about half way there but have recently stumbled across a particular statement (next paragraph).

Let A be an abelian group with a finite number of generators, say {a1,...am}, and assume that A does not contain any element of finite period apart from the unit element. Assume A written additively. Let {a1,...ar} be a maximal subset linearly independent over Z (integers). Let B be the subgroup generated by a1,..,ar. How do I then show that there exists a positive integer d such that dx lies in B for all x in A.

From this point forward, I know how to then prove that A has a basis (I am just stuck on the paragraph above).

Any help on the above would be very appreciated.

Thank you

Answer 1

An idea: for any $\;1\le p\le m-r\;$ , we have that

$$\{a_1,...,a_r,a_{r+p}\}\;\;\text{is $\Bbb Z\,-$ linearly dependent , so there's}\;\; d_p\in\Bbb Z\;\;s.t.$$

$$d_pa_{r+p}\in\text{ Span}_{\Bbb Z}\{a_1,...,a_r\}$$

Define $\;d:=d_1\cdot\ldots\cdot d_p\;$ , so now for any

$$x=\sum_{k=1}^m m_ka_k\in A\;,\;\;m_k\in\Bbb Z\implies $$

$$dx=\overbrace{\sum_{k=1}^r dm_ka_k}^{\in B}+\underbrace{d_2\cdot\ldots\cdot d_p}_{=d/d_1}\cdot \overbrace{d_1a_{r+1}}^{\in B}\, m_{r+1}+\underbrace{d_1d_3\cdot\ldots\cdot d_p}_{=d/d_2}\,\cdot\overbrace{d_2a_{r+2}}^{\in B}m_{r+2}+\ldots\in B$$

Answer 2

By the definition of $B$ (it is a maximal linearly independent subset, where we might have reordered the $a_i$), for $i = 1, \dots, m$ there is a positive integer $d_i$ such that $d_i a_i \in B$. We can take $d_i = 1$ for $i \le r$, and then for $i > r$ such a $d_i$ exists, by maximality.

Let $d$ be the least common multiple of the $d_i$.