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It seems pretty clear to me that both of these are at least uncountable (which I think I could prove with some work). It also seems that you should be able to make some diagonal argument about the two, but I'm not really sure how to make that. I've been trying to think of functions between groups and rings and ways to create groups out of rings and vice-versa, and I even think I've found some injective and/or surjective functions, but I didn't seem to be getting anywhere.

Any suggestions would be great!

Friend-Bob
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4 Answers4

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There are proper-class-many groups and rings - in fact, there are proper class many $X$s, for any reasonable kind of mathematical structure $X$. One way to prove this is the following: given a cardinal $\kappa$, the direct product of $\kappa$-many rings is still a ring, and ditto for groups.

So is there a sense in which there are "more" groups than rings? Certainly every ring is also an abelian group, and there are abelian groups (such as $\mathbb{Q}/\mathbb{Z}$) which cannot be the underlying group of any ring, so in that sense there are "more" groups. But this is like saying there are more rationals than integers.

We can of course say that there are more groups of size $\le n$ than there are rings of size $\le n$, for any fixed finite $n$. This is not trivial: since rings involve two binary operations, there are more possible rings of size $\le n$ than there are possible groups of size $\le n$ (specifically, there are more structures in the language $\{+, -, \times, 0, 1\}$ of size $\le n$ than there are structures in the language $\{+, -, e\}$ of size $\le n$), but it is true.

Once we get to infinite structures, this breaks down completely: there are $2^{\aleph_0}$-many rings of size $\aleph_0$, which is as many as possible. To see this, let $X$ be any set of natural numbers; then we can form a ring by adjoining to $\mathbb{Z}$ the elements ${1\over p_i}$ (where $p_i$ denotes the $i$th prime) for each $i\in X$. This generalizes to show that there are $2^{\kappa}$-many rings of cardinality $\kappa$, for every infinite $\kappa$ (exercise).

NOTE: these aren't just distinct rings, they are rings which are even non-isomorphic!

On the other hand, if we view a countable structure (in a fixed language) as a real number (this can be done in a few ways), then the "probability" that something is either a ring or a group is zero; that is, the reals coding groups, or rings, has Lebesgue measure zero. So they can't be compared meaningfully via measure.

Ben Millwood
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Noah Schweber
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    "There are abelian groups which cannot be the underlying group of any ring" - how so? We can always add the operation $a*b=0$ to any abelian group. – Jonathan Hebert May 20 '15 at 01:18
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    That's not a ring - no unit. – Noah Schweber May 20 '15 at 01:25
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    @user28111: There are plenty of people on both sides of that issue (Wikipedia article on rngs, I understand you may require rings to have units but the OP may not. (I voted +1 on your answer by the way, it's much more in-depth than my own.) – Zev Chonoles May 20 '15 at 01:35
  • That's a fair point - it's hardly an ideal situation. :P – Noah Schweber May 20 '15 at 01:37
  • Coming very very late to this: it's easy to see that all the rings of size $\aleph_0$ are non-isomorphic since the sentence $\exists g (g+g+\cdots+g=1)$ where there are $p_i$ copies of $g$ is clearly true only in those rings where $p_i^{-1}$ has been adjoined, but how do you do this for non-countable $\kappa$? – Steven Stadnicki Jan 05 '17 at 19:09
  • @StevenStadnicki Ah, that was very poor writing on my part: I meant that the result generalizes, not the argument. I don't recall what proof I had in mind when I wrote this, but here's one that I think works: for a Boolean ring $R$, consider the partial order $a\le_Rb$ iff $ab=ba=b$. Now show that for every set $S\subseteq\kappa$, there is a ring $R_S$ of size $\kappa$ such that for all infinite $\alpha<\kappa$ the following are equivalent: (i) $\alpha\in S$, (ii) there is a maximal $\le_{R_S}$-chain of ordertype $\alpha$. – Noah Schweber Jan 05 '17 at 20:49
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The collection of all groups and the collection of all rings are, like the collection of all sets, proper classes (colloquially, they are "too big to be a set").

For any set $S$ whatsoever, you can form the free group $F_S$ and the commutative ring $\mathbb{Z}[S]$, and if you have distinct sets $S\neq T$ then $F_S\neq F_T$ and $\mathbb{Z}[S]\neq\mathbb{Z}[T]$. Thus, there are "at least as many" groups and rings as there are sets.

Zev Chonoles
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    If $S$ and $T$ have the same cardinality, then $F_S\cong F_T$; the answer is still the same, though, if we ask about rings/groups up to isomorphism. – Noah Schweber May 20 '15 at 01:15
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    I feel like we must ask about groups/rings up to isomorphism, because otherwise there is a proper class of trivial groups, which seems fairly silly. – Ben Millwood Nov 17 '15 at 10:38
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Hint: In the setting of Grothendieck Universe, we can talk about the size the of the set of small groups (denoted by Grp) and that of small rings (denoted by Rng), though not very meaningful I think. Then the mapping which sends each small group to the corresponding group ring is a injection from Grp to Rng. I also came up with some injections from Rng to Grp, but none of them is very natural.

Censi LI
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  • Neither "small groups" or "small rings" form a set - in any Grothendieck universe $\mathcal{U}$, both collections will be (internally!) in bijection with all of $\mathcal{U}$. So this doesn't really help distinguish them. – Noah Schweber May 20 '15 at 14:24
  • As for an injection from Rng into Grp: the "underlying group" map can be turned into an injection. Given a ring $R$ with underlying set $S$, we can canonically form a group isomorphic to the additive group of $R$ with underlying set ${(s, {R}): s\in S}.$ Folding in $R$ as a "dummy second coordinate" makes this map injective. – Noah Schweber May 20 '15 at 14:27
  • @user28111 I came up with this injection, but I thought it's a little bit artificial so didn't add it to my answer. By the way, though it has nothing to to with OP's question, I'm curious about whether every proper class of $\mathcal{U}$ is in bijection with $\mathcal{U}$. Do you have any ideal? – Censi LI May 20 '15 at 14:35
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    The answer to your question is: not necessarily! In the language of $ZFC$, which I'm more familiar with, any proper class will surject definably onto the class of ordinals, $ON$, but $ON$ may not surject definably onto all of $V$. And of course the same situation can internally to a universe. Having a definable surjection of $ON$ onto $V$ is the same as saying that the whole universe is well-ordered - this is an axiom of global choice. (cont'd) – Noah Schweber May 20 '15 at 16:06
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    As an example where $ON$ is strictly smaller than $V$ (in that $V$ surjects onto $ON$, but not vice-versa), see Joel David Hamkins' answer to http://mathoverflow.net/questions/110799/does-zfc-prove-the-universe-is-linearly-orderable. He shows there is a model of set theory which has no definable linear order. – Noah Schweber May 20 '15 at 16:07
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    Why are rings and groups in bijection with $V$? Well, it's basically because of the "free ???" construction: given any set $X$, I have a definable way of forming a group/ring/??? $F(X)$, and my construction $F$ is injective. This gives me an injection of $V$ into groups/rings/???, which can be massaged into a bijection with some effort. Note that for the class of ordinals $ON$, there is no "free ordinal" construction, and of course we can't even get a surjection of $ON$ onto $V$ (much weaker than an injection of $V$ into $ON$). – Noah Schweber May 20 '15 at 16:10
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    @user28111 Thank you very much. – Censi LI May 20 '15 at 16:16
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There are more groups than rings in the sense that the forgetful functor from Rings to Groups which selects the additive group of a ring is not surjective, even if you restrict it to Abelian Groups. In other words, not every abelian is the additive group of a ring.

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