2

There are many simple examples of groups with infinite cardinality which are known outside of set theory, such as $\mathbb Z$, which has cardinality $\aleph_0$, and $\mathbb R$, which has cardinality $2^{\aleph_0}$. Assuming the continuum hypothesis, $\mathbb R$ is also a familiar example of a group with cardinality $\aleph_1$. However, is there a relatively simple example of a group where we can prove, in $\mathsf{ZFC}$ alone, that it has cardinality $\aleph_1$?

Joe
  • 19,636
  • 7
    The free group on $\aleph_1$-many generators? – Noah Schweber Nov 27 '23 at 00:16
  • @NoahSchweber, :) :) But, yes, very true. :) – paul garrett Nov 27 '23 at 00:18
  • @NoahSchweber: I'm not overly familiar with the free group construction, since I haven't yet got to that part of the algebra book I'm reading. However, I suppose that example is about as simple as one could hope for, so you could post it as an answer, if you wish. – Joe Nov 27 '23 at 00:18
  • Sure, of any infinite cardinality. See https://math.stackexchange.com/questions/1296889/fields-of-arbitrary-cardinality for a stronger statement. – Martin Brandenburg Nov 27 '23 at 00:25
  • See also https://math.stackexchange.com/questions/105433/does-every-set-have-a-group-structure. So the question here is a duplicate. – Martin Brandenburg Nov 27 '23 at 00:27
  • 3
    @Joe -- a possibly more familiar version of Noah's answer is any $\mathbb{Q}$-vector space with $\aleph_1$ many basis vectors. In fact, an $\aleph_1$-dimensional vector space over any finite field would also work. – HallaSurvivor Nov 27 '23 at 00:27
  • @MartinBrandenburg: Hmm. On the one hand, the examples you have provided are very useful. On the other hand, I'm curious to see if there are any non-abelian examples. (But perhaps Noah's example suffices.) – Joe Nov 27 '23 at 00:28
  • 1
    Sure, take any abelian example and take the product with $S_3$ or any finite non-abelian group. – Martin Brandenburg Nov 27 '23 at 00:29
  • @MartinBrandenburg: Ah, I see. I've marked my post as a duplicate. Thanks to all for your help. – Joe Nov 27 '23 at 00:31
  • @MartinBrandenburg: Would it be possible for you to add your second linked post to the duplicates list, too? – Joe Nov 27 '23 at 00:35
  • 1
    By the way none of the above constructions require choice - $\aleph_1$ is nice and well-orderable so facts about cardinal arithmetic are true "for free". The example Wikipedia gives is the set of finite subsets of a set of cardinality $\aleph_1$ under symmetric difference, which is secretly an instance of HallaSurvivor's remark "an $\aleph_1$-dimensional vector space over any finite field". – Izaak van Dongen Nov 27 '23 at 00:42
  • I tried, but for some reason I can't this time. The option "Edit" is not available. @Joe – Martin Brandenburg Nov 27 '23 at 00:42
  • @IzaakvanDongen: Thank you for your added remarks. That seems like the simplest example to my mind, but perhaps I am just showing my ignorance of both group theory and set theory :) – Joe Nov 27 '23 at 00:48

0 Answers0