9
$\begingroup$

A group $G$ is locally cyclic if whenever $H \le G$ is a finitely generated subgroup then $H$ is cyclic. If $G$ is a locally cyclic group then $G$ is isomorphic to a quotient of a subgroup of the rational numbers under addition. An online proof of this fact appears at groupprops. However, despite some searching, I have been unable to find a proof in the published literature on abelian groups.

Is there a published paper or textbook that has a proof that every locally cyclic group is isomorphic to a quotient of a subgroup of the rational numbers?

$\endgroup$

1 Answer 1

9
$\begingroup$

For torsion-free groups it is proved in Kurosh, "Group theory", See 3d edition, Chapter VIII, Section 30 (of course the result can be found in the 1st edition as well). Oroginally it was proved in Reinhold Baer, "Abelian groups without elements of finite order". Duke Math J. 3 (1): 68–122, 1937. For groups with torsion, it follows from old results as well but I am not sure anybody specifically mentioned it somewhere. Of course you need to look at Fuchs, "Infinite Abelian groups" (both volumes). If it is not there, it is probably not anywhere else.

$\endgroup$
2
  • $\begingroup$ Many thanks for these references. For torsion-free groups the result is mentioned on page 109 of Fuchs "Infinite Abelian Groups", Volume II. I cannot find the torsion case in Fuchs in any explicit form. Marshall Hall's "Theory of Groups" page 340 has "The additive group of rationals $R_+$ is a locally cyclic group which is aperiodic, and the group $R_+$ modulo $1$ is a periodic locally cyclic group. It is not too difficult to show that a locally cyclic group is a subgroup of one of these two groups." (No proof is given.) $\endgroup$ Oct 4, 2011 at 17:20
  • $\begingroup$ @Mark: The proof should go like this. First embed your Abelian group into a full group. Note that the full group can be assumed locally cyclic too. Then use the description of Abelian full groups. $\endgroup$
    – user6976
    Oct 4, 2011 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.