# Finitely generated abelian group has a regular normal form

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
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.
• 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.