Are there more groups than rings?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.