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Let $R$ be a ring. An element $r \in R$ is a unit of $R$ if there is some $s \in R$ such that $rs=sr=1$. Denote the set of units of $R$ by $R^{\times}$. It is easy to verify that $R^{\times}$ is a group under the multiplicative operation of $R$.

I am trying to find a ring $R$ such that $R^{\times} \cong D_4$. Recall that $D_4$ is the dihedral group of order 8 which has the following presentation: $\langle a,b ~|~ a^4 = 1, b^2 = 1, bab=a^3 \rangle$.

More generally, is every finite group the group of units of some ring?

Liam
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    @DietrichBurde: I think this is not a duplicate. – mesel Mar 31 '15 at 21:10
  • Not a duplicate. – Jim Mar 31 '15 at 21:14
  • I feel that my first question was not a duplicate, but then my generalized question was a duplicate. The linked question's answer is focused on where $R^{\times}$ are cyclic groups, but I think it is good that it is marked as a related question. – Liam Apr 01 '15 at 19:11

2 Answers2

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Yes, note that $\text{Aut}(C_4\times C_2)\simeq D_4$ ($C_n$ denotes the cyclic group of order $n$). So $D_4\simeq R^\times$ for $R=\text{End}(C_4\times C_2)$.

Censi LI
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It is not true that every finite group is the group of units of some ring:

Suppose that $R$ is a ring with $R^\times \cong C_5$, and let $t\in R^\times$ be a generator. By passing to the ring generated by $1$ and $t$, we may assume that $R$ is a quotient of $\mathbb{Z}[X]/(X^5-1)$.

We must have $-1=1$, because $R^\times$ has no elements of even order, so $R$ is further a quotient of $\mathbb{F}_2 [X]/(X^5-1)$. But this ring is isomorphic to $\mathbb{F}_2 \times \mathbb{F}_{16}$, and we can check that its four quotients have groups of units with sizes $1$ and $15$ only.

Andrew Dudzik
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