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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?

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    $\begingroup$ See also math.stackexchange.com/questions/384422/… $\endgroup$
    – lhf
    Sep 24, 2019 at 0:27
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    $\begingroup$ Jack Schmidt's answer contradicts your claim about f.g. abelian groups. $\endgroup$ Sep 24, 2019 at 0:29
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    $\begingroup$ 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. $\endgroup$ Sep 24, 2019 at 0:30
  • $\begingroup$ 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. $\endgroup$
    – quid
    Sep 24, 2019 at 13:57

3 Answers 3

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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.

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  • $\begingroup$ 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.) $\endgroup$
    – quid
    Sep 24, 2019 at 13:50
  • $\begingroup$ @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. $\endgroup$ Sep 24, 2019 at 17:29
  • $\begingroup$ A proof for C_5 was the motivation of the other question, see math.stackexchange.com/questions/384362/… $\endgroup$
    – quid
    Sep 24, 2019 at 17:35
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    $\begingroup$ Ah, I didn’t see that one! Welp. $\endgroup$ Sep 24, 2019 at 17:41
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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.

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    $\begingroup$ 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 $GL_{n_i}(\mathbb{F}_{2^{k_i}})$. $\endgroup$ Sep 24, 2019 at 0:27
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    $\begingroup$ 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). $\endgroup$ Sep 24, 2019 at 2:16
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Kenta S
    May 21, 2021 at 15:02
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    $\begingroup$ @KentaS Right. Also, the functor is neither full nor faithful. $\endgroup$
    – user403337
    May 21, 2021 at 20:54

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