# Is every group the unit group of some ring?

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?

• – lhf Sep 24 '19 at 0:27
• Your claim that all finitely generated abelian groups are R-groups is false, since not all cyclic groups are R-groups. For example, $\mathbb{Z}_5$ is not the group of units of any ring. – diracdeltafunk Sep 24 '19 at 0:30
• The question is essentially a duplicate of the linked to one; the linked to one is only for finite groups, but the negative answer there provides a negative answer here. The partial progress you mention seems false. If you want to ask about what is known in the positive direction for infinite groups, I recommend you ask this anew mentioning the older question (not this one). Or focus this question down to the F_2[G] aspect. – quid Sep 24 '19 at 13:57

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 $$\mathbb{F}_3, \mathbb{F}_4, \mathbb{F}_5$$.)

• How was this not a duplicate? (Maybe due to finite vs infinite yet your answer says nothing for the infinite case. Or even if one focuses on the F_2[G] aspect as the other answer, but you didn't do that either.) – quid Sep 24 '19 at 13:50
• @quid: I suppose it is a duplicate. Jack’s answer to the other question includes no proofs and links to papers which aren’t freely available, so I thought it would be nice to have a proof of something at least. I focus on elements of finite order here but this argument rules out many infinite groups. – Qiaochu Yuan Sep 24 '19 at 17:29
• A proof for C_5 was the motivation of the other question, see math.stackexchange.com/questions/384362/… – quid Sep 24 '19 at 17:35
• Ah, I didn’t see that one! Welp. – Qiaochu Yuan Sep 24 '19 at 17:41

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.

• More generally, if $|G|$ is finite and odd then $\mathbb{F}_2[G]$ is semisimple by Maschke's theorem, so is isomorphic to a product of matrix algebras $M_{n_i}(\mathbb{F}_{2^{k_i}})$ and has group of units a product of general linear groups $GL_{n_i}(\mathbb{F}_{2^{k_i}})$. – Qiaochu Yuan Sep 24 '19 at 0:27
• In fact we have $\mathbb{F}_2[C_5] \cong \mathbb{F}_2 \times \mathbb{F}_{16}$, and the the element you've written down is an element of order $3$ in $\mathbb{F}_{16}^{\times}$ (you can verify this by squaring it and checking that its square equals its inverse). – Qiaochu Yuan Sep 24 '19 at 2:16

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.

• So, in particular, the functor $\textbf{Grp}\to\textbf{Ring}\cong\mathbb Z-\textbf{Alg}$ sending $G$ to $\mathbb Z[G]$ is left-adjoint to the group of units functor. – Kenta S May 21 at 15:02
• @KentaS Right. Also, the functor is neither full nor faithful. – user403337 May 21 at 20:54