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$\begingroup$

I have only recently encountered algebraic number theory and was wondering if this is the case. If the answer to the question is yes, then can we explicitly construct the domain $D$ ?

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    $G$ must be abelian...2012-01-24
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    I know this is a repeat... Yes: [here it is](http://math.stackexchange.com/questions/10949/finite-abelian-groups-as-class-groups) for finite abelian groups.2012-01-24
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    @Fortuon: It should be obvious from the definition of ideal class group that it is abelian. If this is not obvious, perhaps you should first learn what the ideal class group is!2012-01-24
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    @ZhenLin: My apologies for the notation abuse. I have corrected it.2012-01-24
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    I don't understand your edit. The ideal class group is commutative because multiplication is commutative in $\mathcal{O}_K$, and any reasonable definition of fractional ideal classes of an integral domain $D$ will make use of multiplication in $D$. Your question doesn't make sense for general non-commutative domains because there isn't even a reasonable definition of fraction field in that case.2012-01-24
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    The paper by Claborn cited in @Mariano's answer does not seem to be restricted to *finite* groups.2012-01-27
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    If you feel that @Mariano has answered your question fully then just accept his answer.2012-01-27

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See Claborn, Luther (1966), "Every abelian group is a class group", Pacific J. Math. 18: 219–222. You can get it here

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    Our dear Pete Clark has a recent result on this subject. You should find it in his webpage.2012-01-24
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    Have you looked?2012-01-24