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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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    $\begingroup$ What is your definition of "regular normal form"? $\endgroup$
    – Arturo Magidin
    Mar 25, 2012 at 3:40
  • $\begingroup$ 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. $\endgroup$
    – Miranda
    Mar 25, 2012 at 3:51
  • $\begingroup$ Are you familiar with the structure theorem for finitely generated abelian groups? $\endgroup$
    – Arturo Magidin
    Mar 25, 2012 at 3:55
  • $\begingroup$ I am drawing a blank. I vaguely recall the notion of PIDS, but that is it. $\endgroup$
    – Miranda
    Mar 25, 2012 at 3:58
  • $\begingroup$ I don't know what PIDS stands for, sorry. For the structure theorem, you can see Wikipedia. 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. $\endgroup$
    – Arturo Magidin
    Mar 25, 2012 at 4:05

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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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  • $\begingroup$ thank you for your help. I have seriously forgotten my hungerford. I guess I need to break it out again. Thanks again. $\endgroup$
    – Miranda
    Mar 25, 2012 at 4:09

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