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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"? –  Arturo Magidin Mar 25 '12 at 3:40
    
Please do not use [homework] as the only tag for your questions. –  Arturo Magidin Mar 25 '12 at 3:40
    
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. –  Miranda Mar 25 '12 at 3:51
    
Are you familiar with the structure theorem for finitely generated abelian groups? –  Arturo Magidin Mar 25 '12 at 3:55
    
I am drawing a blank. I vaguely recall the notion of PIDS, but that is it. –  Miranda Mar 25 '12 at 3:58

1 Answer 1

If $G$ and $H$ are two groups having regular normal forms, then you should have little problem showing that the direct product $G\times H$ also has a regular normal form.

Now a finitely generated abelian group is a direct product of finitely many cyclic groups —this is part of the content of the structure theorem that Arturo mentioned in the comments— so the above observation allows us to reduce our consideration to cyclic groups.

  • If a cyclic group $G$ is finite of order $n$ and $g$ is a generator, then $G=\{g^0,g^1,\dots,g^{n-1}\}$. The restriction of the canonical map $\{g,g^{-1}\}^*\to G$ to the finite subset $\{\varepsilon,g,g^2,\dots,g^{n-1}\}$ of its domain, which is of course regular, is a bijection.

  • On the other hand, if $G$ is cyclic and infinite, let $t$ be a generator and let $=t^{-1}$ be its inverse. The restriction of the canonical map $\{t,s\}^*\to G$ to the language denoted by the regular expression $t^*\cup ss^*$ is a bijection.

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thank you for your help. I have seriously forgotten my hungerford. I guess I need to break it out again. Thanks again. –  Miranda Mar 25 '12 at 4:09

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