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Ring with Z as its group of units?

Given a group $G$, does there always exist a ring $R$ such that $R^\times \cong G$? I feel like this isn't true but that's just a hunch. One of my friends asked me this yesterday and we couldn't come up with an answer so I thought I'd ask here.

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marked as duplicate by Alex Bartel, Anton Geraschenko Oct 19 '11 at 1:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Does the group ring $F_2[G]$ work? –  George Lowther Oct 19 '11 at 0:18
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This came up before. See mathoverflow.net/questions/75192/… –  Faisal Oct 19 '11 at 0:24
    
And Jesse Elliot's comment says that $F_2[G]$ does not work. –  George Lowther Oct 19 '11 at 0:27
    
When $|G|$ is finite and odd, then it must be the direct product of cyclic groups of order $2^k-1$ for various $k$; in other words, you only have to worry about the groups coming from $R=GF(2^{n_1})\times GF(2^{n_2})\times\cdots$. –  Steve D Oct 19 '11 at 0:45
    
Nice question! –  Anton Geraschenko Oct 19 '11 at 1:21
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