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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

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    Asaf Karagila answered that in a comment to your previous question. The answer is yes, and it was proved by W.R. Scott in a paper in 1952. – Derek Holt Apr 09 '15 at 14:47
  • @DerekHolt : Yes, true , but Asaf Karagila actually answered in contrapositive :-) , I am wondering whether there is a proof , not too heavy , of this direct claim using choice ... –  Apr 09 '15 at 14:52
  • But that's a different question. The answer to the question you asked is yes. – Derek Holt Apr 09 '15 at 14:53
  • @DerekHolt : I have edited to make it clear –  Apr 09 '15 at 14:55
  • If the proof there doesn't use the axiom of choice, then of course assuming the axiom of choice will not damage the proof. But the proof does use the axiom of choice, since otherwise it is consistent that a counterexample exists. – Asaf Karagila Apr 09 '15 at 14:57
  • @AsafKaragila : Yes , yes , I get it , but I am asking for a direct proof here , not the other way around ( that to show it fails under not assuming A.C. ) , shouldn't it be easier ? –  Apr 09 '15 at 14:58
  • I didn't answer it in a contrapositive way. At all. I merely mentioned that the axiom of choice is in fact essential to the proof. I then referred you to a paper which includes a proof. – Asaf Karagila Apr 09 '15 at 15:06

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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".

Asaf Karagila
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