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(Stacked bases theorem) If $A$ is a finitely generated free abelian group, and $B$ is a subgroup of $A,$ of the same rank, say $m$, then there is a basis $\{a_{i}: 1 \leq i \leq m \}$ of $A$ and there are elements $d_{1},d_{2},\ldots d_{m}$ of $R$ such that $d_{i} | d_{i+1}$ for each $i$ and $\{ d_{i}a_{i}: 1 \leq i \leq m \}$ is a generating set for $B$.

Some books prove the stacked bases theorem before the structure theorem for finitely generated abelian group. But I wonder how to do it conversely.

If we assume the structure theorem for finite abelian group which shows that $A/B$ is a direct sum of at most $m$ cyclic groups. How to use this fact to show the stacked bases theorem above?

Shaun
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No One
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  • Let me point out that, leaving the condition $d_i\mid d_{i+1}$ aside, your question is essentially what allegedly Marcus asks for in his book "Number Fields" (exercise 27.(b) in the chapter 2). I say "allegedly" because the exercise states "there is this theorem which says finite abelian groups are products of cyclic groups. Deduce stacked bases theorem" with no indication on how one might use it. – Wojowu Oct 12 '16 at 15:15
  • Possibly the easiest way to see this is to write down a $ \mathbf Z $-linear map $ A \to A $ whose image is $ B $, and put it in Smith normal form. The structure theorem also drops out immediately from this idea. – Ege Erdil Oct 12 '16 at 15:25
  • @Starfall Although this idea works, I suppose that OP's intent is to try and get a proof directly from the structure theorem without having to use other, intricate or not, ideas. – Wojowu Oct 12 '16 at 15:27

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