Can a group have $\aleph_1$ many subgroups?

Question

I know that any countable group has either countably many or continuum many subgroups (source), but I'm curious about uncountable groups.

It seems like the proof for countable groups $G$ crucially uses the fact that $2^G$ is a polish space, which I think isn't true anymore when $G$ is uncountable. So then:

Is it possible for an uncountable group to have exactly $\aleph_1$-many subgroups? (Assuming $\lnot \mathsf{CH}$, of course)

More generally, for uncountable cardinals $\kappa$ and $\lambda$, is it possible to have a group of size $\kappa$ with exactly $\lambda$ many subgroups? There are definitely some easy combinations of $\kappa$ and $\lambda$ that we can rule out, but this problem seems like it is probably hard.

Thanks!

Answer 1

As a partial answer, let me prove that any uncountable abelian group $A$ always has $2^{|A|}$ subgroups. Let $T$ be the torsion subgroup of $A$. Then $|A|=|T|\times|A/T|$ so at least one of $T$ and $A/T$ must have the same cardinality as $A$. Thus we may assume either $A$ is torsion or $A$ is torsion-free.

If $A$ is torsion, then $A$ is an extension of a direct sum of cyclic groups $B$ by a divisible group $C$. At least one of $B$ and $C$ has the same cardinality as $A$, and each of them is a direct sum of countable groups. So at least one of $B$ and $C$ is a direct sum of $|A|$ nontrivial groups and by taking arbitrary subsets of the summands we can get $2^{|A|}$ different subgroups.

If $A$ is torsion-free, then we can pick a basis for the $\mathbb{Q}$-vector space $A\otimes\mathbb{Q}$ which consists of elements of $A$. This gives a subgroup of $A$ which is a free abelian group on $|A|$ generators, and again by taking arbitrary subsets of the generators we can get $2^{|A|}$ different subgroups.

Unfortunately, the nonabelian case seems much harder.

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On the other hand, let me point out that it is possible to have a finitary algebraic structure with exactly $\aleph_1$ substructures (so any proof that this is impossible for groups will have to use something more special about groups). For instance, consider $\omega_1$ equipped with the following binary operation $f$. For each infinite $\alpha<\omega_1$, fix a bijection $f_\alpha:\omega\to\alpha$. Now define $f$ as follows:

• If $n$ is finite, then $f(0,n)=n+1$.
• If $n$ is finite and $\alpha$ is infinite, then $f(n,\alpha)=f_\alpha(n)$.
• In all other cases, $f(\alpha,\beta)=0$.

The first and third bullet points imply that any nonempty substructure $X\subseteq\omega_1$ must contain all finite ordinals. Then the second implies that if $\alpha\in X$ then $\alpha\subseteq X$, so any substructure must itself be an ordinal. Conversely, any infinite ordinal $\alpha\leq\omega_1$ is easily seen to be a substructure. So, the nonempty substructures of $\omega_1$ are exactly the infinite ordinals $\leq\omega_1$, and in particular there are exactly $\aleph_1$ of them.

Answer 2

Yes, at least assuming the continuum hypothesis. In the following paper, Shelah proves that there is a group of size $\aleph_1$ whose proper subgroups are all countable.

Shelah, Saharon, On a problem of Kurosh, Jonsson groups, and applications, Word problems II, Stud. Logic Found. Math. Vol. 95, 373-394 (1980). ZBL0438.20025.

Since every proper subgroup is countable, there are at most $\aleph_1^{\aleph_0}=2^{\aleph_0}$ subgroups. It is not hard to show there are at least $\aleph_1$ subgroups, so under CH there are exactly $\aleph_1$ subgroups.