How many non-isomorphic algebraic structures (i.e. magmas, monoids, groups etc.) are there with countably infinite order?

Question

For structures of finite order it seems obvious to me that there are countably infinite in total, by a simple diagonalization argument (starting at all of order 1, then 2 etc.). It is however not clear to me for structures of countably infinite order how many there will be in total. Intuitively it seems very unlikely to also be countably infinite, but I'm not even sure how to show just that.

Answer 1

For most types of structure naturally encountered - including groups, rings, fields, magmas, semigroups, Boolean algebras, and so forth - there are continuum many (= as many as possible) countable examples up to isomorphism.

However, there are certain types of structure where the situation is different. For example, up to isomorphism there are only countably many countable vector spaces over $\mathbb{Q}$, while there are no countable vector spaces over $\mathbb{R}$ whatsoever (fine, except the trivial vector space $\{0\}$).

Additionally, there are types of structure where the answer to the question is at least to a certain extent independent of the usual axioms of set theory. For example, up to isomorphism there are $\aleph_1$-many countable well-orderings. Are there continuum-many? Well, that depends.

So there are some general heuristics, but the situation really depends on the specific type of structure being considered.

Answer 2

Let me turn my comment to @NoahSchweber's answer into a quick proof that there are uncountably many countable (abelian) groups up to isomorphism. This therefore implies that there are uncountably many such monoids and magmas too, as groups are additionally monoids and magmas.

If $\mathcal{P}$ is a set of primes then let $D_{\mathcal{P}}$ be the direct sum of cyclic groups of order $p\in\mathcal{P}$. The groups $D_{\mathcal{P}}$ are

• countable (as they are the direct sums of finite groups), and
• pairwise non-isomorphic (as if $p\in\mathcal{P}$ but $p\not\in\mathcal{P}'$ then $D_{\mathcal{P}}$ has an element of order $p$ but $D_{\mathcal{P}'}$ does not).

Finally, there are uncountably many sets $\mathcal{P}$ (why?), and so the claim holds: there are uncountably many countable groups up to isomorphism.

A similar idea works in lots of other categories, e.g. for rings take the direct sum of the rings $\mathbb{Z}/p\mathbb{Z}$, $p\in\mathcal{P}$.

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The structures $D_{\mathcal{P}}$ can be turned into "strict" monoids, magmas etc. by taking the direct sum with a "strict" monoid or magma. For example, writing $\mathbb{N}$ for the non-zero natural numbers and $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$, we could define:

• $D_{\mathcal{P}}^{\operatorname{semigroup}}:=D_{\mathcal{P}}\oplus\mathbb{N}$,
• $D_{\mathcal{P}}^{\operatorname{monoid}}:=D_{\mathcal{P}}\oplus\mathbb{N}_0$,
• $D_{\mathcal{P}}^{\operatorname{magma}}:=D_{\mathcal{P}}\oplus M$ where $M$ is this thing.

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Finally, asking for groups etc. with specific properties may change you answer. In particular, in group theory we often talk about "finitely generated" and "finitely presentable" groups, and such groups are necessarily countable, but while there are uncountably many finitely generated groups (in fact, B.H. Neumann proved in 1937 that there are uncountably many $2$-generated groups), there are only countably many finitely presentable groups - this is essentially because there are only finitely many finite presentations, and because a presentation defines the group (much like a Cayley table does, but finite presentations still allow for infinite structures).