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Prove that every finitely generated abelian group admits a regular normal form. I am having some trouble getting my head wrapped around this problem. If anyone can offer suggestions or help it would be greatly appreciated.

Added. Given a group, say $G$, and a generating set $S$, a normal form is a subset of the free monoid $\{S\cup S^{−1}\}$. This maps bijectively to $G$ under the evaulation map $\alpha \colon \{S \cup S^{-1}\}^{*} \to G$. Then a normal form say $\mathrm{NF} \subseteq \{S \cup S^{-1}\}$, which will be thought of as a language, we just want it to be a regular language.

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    What is your definition of "regular normal form"?2012-03-25
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    Given a group say G and a generating set S, a normal form is a subset of the free monoid $\{S \cup S^{-1}\}$. This maps bijectively to G under the evaulation map $\alpha$ : $\{S \cup S^{-1}\}$^{*} $\rightarrow$ G. Then a normal form say NF $\subset$ $\{S \cup S^{-1}\}$, which will be thought of as a language, we just want it to be a regular language.2012-03-25
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    Are you familiar with the structure theorem for finitely generated abelian groups?2012-03-25
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    I am drawing a blank. I vaguely recall the notion of PIDS, but that is it.2012-03-25
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    I don't know what PIDS stands for, sorry. For the structure theorem, you can see [Wikipedia](http://en.wikipedia.org/wiki/Fundamental_theorem_of_finitely_generated_abelian_groups#Classification). I don't know if the theorem will give you a "normal form" in the sense you require (I don't quite remember what "regular language" means), but it seems like a natural place to start.2012-03-25
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    PIDS, are just Principal Ideal Domains. Thanks so much for the help.2012-03-25
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    Are you required to prove that they admit a regular normal form with respect to an arbitrary finite generating set, or are you allowed to choose the generating set yourself? In the second case, as Mariano Suárez-Alvarez indicated in his answer, just express the group as a direct product of cyclic groups, and choose generators of the cyclic direct factors. It is much harder to do it with an arbitrary finite generating set, but the result is still true.2012-03-25

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