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What is known about the groups G for wich there exist a unitary ring R, such that $R^{\times} \simeq G$? I can easily prove that

The only G cyclic with this property(Edit:and odd order) are those who factors via pairwise coprime integers of the form $2^n-1$. But i'm wondering for more general group(also restricting just to abelians) what is known and what would be interesting to try to find out.

Edit:sorry i meant the only G cyclic, with odd cardinality, i forgot the even ones. See comment

user91880
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  • See http://math.stackexchange.com/questions/384422/which-finite-groups-are-the-group-of-units-of-some-ring for counterexamples (such as any finite field of odd characteristic). – Jack Schmidt Feb 12 '14 at 19:59
  • Yes you're right i forgot the even ones. But as i can see from that paper of Pearson and Schneider the even part is right. So my question is wich groups are unkown today(if they are unity group of some ring) and it could be a good idea to try to work out. Especially wich abelian group. – user91880 Feb 12 '14 at 20:49
  • Anyway thanks for the reference(i guess in even order things get more complicated because my argument in odd case is very simple and is due to the fact that -1 should be 1, so you can pass mod 2 and taking $\mathbb{Z}[g]$ where g is a generator you indeed get $\mathbb{F}_2[g]$ that factors as product of local rings, and if you want odd cardinality they have to be fields. And from this everything is under control. – user91880 Feb 12 '14 at 20:52

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