Is every group the unit group of some ring?

Question

Let the functor $F\colon\bf Ring\rightarrow\bf Grp$ send the ring $A$ to its group of units $A^\times,$ and the ring homomorphism $f\colon A\rightarrow B$ to the group homomorphism $f^\times\colon A^\times\rightarrow B^\times:a\mapsto f(a)$.

I was curious about this functor, and in particular, whether it is essentially surjective. That is, for any group $G$ (not just finite,) is there a ring $A$ such that $G\cong A^\times$? If not, what groups $G$ satisfies this? A similar question was asked in this question, but what can be said for infinite groups, or groups in general? Calling groups that satisfy this condition R-groups, I have proved that any finitely generated abelian group is an R-group.

I have no idea what to do from now, but I have a conjecture that the unit group of the group ring $\mathbb F_2[G]$ is isomorphic to $G.$ If this is true, certainly, all groups are R-groups, and the functor $F$ is essentially surjective, but I am having trouble proving it. Can anyone help me?

Answer 1

No, this is already false for finite abelian groups.

A ring either has characteristic $2$ or it has a non-identity unit $-1$ which is central of order $2$, so if a group $G$ doesn't have such an element then it can only arise as the group of units of a ring of characteristic $2$.

Let $R$ be such a ring and consider an element $r \in R^{\times}$ of odd prime order $p$. (Edit: There was a sloppy argument here with an error which has now been corrected, twice!) It generates a subring of $R$ given by some quotient of the group algebra $\mathbb{F}_2[C_p]$ in which $C_p$ embeds. By Maschke's theorem $\mathbb{F}_2[C_p]$ is semisimple and hence a finite product of finite fields $\mathbb{F}_{2^k}$, and $C_p$ embeds into some $\mathbb{F}_{2^k}$ iff $p | 2^k - 1$.

So $R^{\times}$ has an element of order $2^k - 1$ where $k$ satisfies $p | 2^k - 1$. Hence:

Any group $G$ which

1. does not have a central element of order $2$ and
2. has an element of odd prime order $p$ but does not have an element of order $2^k - 1$ satisfying $p | 2^k - 1$

is not the group of units of a ring.

The smallest such group is the cyclic group $C_5$ (mentioned by diracdeltafunk in the comments), which has odd order and hence no elements of order $2$, and which has an element of order $5$, but does not have an element of order $2^4 - 1 = 15$ or larger. (And the cyclic groups $C_2, C_3, C_4$ are the groups of units of the finite fields $\mathbb{F}_3, \mathbb{F}_4, \mathbb{F}_5$.)

See also the classification described by Jack Schmidt in the answer linked to by lhf in the comments.

Answer 2

Your statement about $\mathbb{F}_2[G]$ is incorrect. Consider when $G = \mathbb{Z}_5$, generated by some element $a$ with $a^5 = e$. Then,

$$(e + a^2 + a^3)(e + a + a^4) = (e + a^2 + a^3) + (a + a^3 + a^4) + (a^4 + a + a^2) = e + (a+a) + (a^2+a^2) + (a^3+a^3) + (a^4+a^4) = e$$

So, the unit group of $\mathbb{F}_2[G]$ includes the natural inclusion of $G$, but it also includes $e + a^2 + a^3$, as shown above.

For reference on how I found this example: I can call the "weight" of an element in $\mathbb{F}_2[G]$ the number of nonzero coefficients, so both of the elements above have weight 3, while $e+a$ has weight 2. Clearly weights multiply, so if we want to end up with an odd weight element like $e$, we must start with two odd-weight elements; and we don't want to use elements of weight 1. So we need $|G|$ at least 3. With $G = \mathbb{Z}_3$, there is only one element with odd weight more than 1, and it doesn't square to $e$. So I jumped to $\mathbb{Z}_5$ and it worked.

Answer 3

The group of units functor is not "surjective".

We do have, however, an adjunction, which is the nearest thing to a categorical inverse, by considering the group ring associated to a given group. $\mathcal R[G] $, informally, consists in all the linear combinations of elements of $G $, weighted by elements of the ring $\mathcal R $. Thus we get a functor from $\bf {Grp} $ to the category $\bf {\mathcal R-Alg} $ of $\mathcal R $-algebras.