# A basic question about rings

Perhaps this is a trivial question, but I have no idea how to justify it.

Call a pair of groups $(G_1, G_2)$ ring-compatible if $G_1$ is abelian and there exists a ring $R$ with addition and multiplication $(+, \ast)$ say satisfying the condition that $(R, +)$ is isomorphic to $G_1$ and the group of units of $R$, say $U(R)$, satisfy $(U(R), \ast) \cong G_2$.

For instance, the existence of finite fields of order $p^n$ for a prime $p$ implies that the pair $(G_1, G_2)$ with $G_1$ being the group $(C_p)^n = C_p \times \cdots \times C_p$ and $G_2$ being the cyclic group of order $p^n - 1$ is a ring-compatible pair.

Are all pairs of groups $(G_1, G_2)$ with $G_1$ abelian and if $G_1, G_2$ are finite groups, $|G_1| > |G_2$, ring compatible?

I apologize in advance if this question is trivial.

• Yes that is a useful remark; thank you! Commented Jan 17, 2014 at 16:59
• But not even every cyclic group is the unit group of some ring. Commented Jan 17, 2014 at 17:01
• Also, if $G_1$ is or prime order, then there is only one choice for $R$, and thus for $G_2$. Commented Jan 17, 2014 at 17:03
• Let's extend this question to infinite group: for instance Is the pair $(\mathbb{C}, \mathbb{R})$ ring compatible? Commented Jan 17, 2014 at 23:08
• According to Kadison conjecture a related question can be as follows: Let G be a torsion free group, Is $\mathbb{C}G, G$ ring compatible? Commented Jan 17, 2014 at 23:11

## 2 Answers

The answer for your question is NO, not every pair of groups is ring compatible, even if the first group is abelian. There are abelian groups $G_1$ (called Nil groups), in which only zero multiplication turns $G_1$ into a ring. A complete discussion of your question can be find in "L. Fuchs, Infinite abelian groups" (Despite the name of book, you can find facts about additive group of finite rings)

• As I remember, two former Ph.D student of our faculty did some works on additive groups of rings and their characterizations (I was the advisor of them) and you may find some of their papers by searching: A. Najafizadeh and F. Karimi. Commented Jan 18, 2014 at 3:30

Note that the second group relates to (and sometimes embeds in) the endomorphism algebra of the first group (this is for the multiplicative semigroup as a whole; the group of units has an obvious correspondence with a subgroup of automorphisms of the first group), so you will have some obvious limitations. I suspect the answer you want lies in the study of modules .

Gerhard "Maybe Another Question Is Meant?" Paseman, 2014.01.17