Assuming the axiom of choice , how to prove that every uncountable abelian group must have an uncountable proper subgroup ? Related to Does there exist any uncountable group , every proper subgroup of which is countable? , Asaf Karagila answered it there in a comment , but in a contrapositive way , I am looking for a direct proof of this claim assuming choice . Thanks in advance
Yes. Assuming the axiom of choice the answer is positive. You can find the proof in W.R. Scott's paper:
Scott, W. R. "Groups and cardinal numbers." Amer. J. Math. 74, (1952). 187-197.
The axiom of choice is used there for all manner of cardinal arithmetics.
Without the axiom of choice it is no longer necessary that the proof can go through. Because it is consistent that there is a vector space over $\Bbb Q$ which is not finitely generated, but every proper subspace is in fact finitely generated.
Of course this means that the vector space is not countable, since otherwise the usual arguments would show it has a countable basis, and therefore it has an infinitely dimensional proper subspace.
The result is originally due to Lauchli, and in my masters thesis I refurbished the argument and showed that you can also require "countably generated" instead "finitely generated".