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December 20

A subgroup of the Baer-Specker group

Let G be the Baer Specker group. If we call H the subgroup that consists of all bounded sequences what’s H’s usual name in the literature and what properties does it have? Such as: does it still fail to be free abelian? Is H invariant under all automorphisms of G? Is there a subgroup akin to functions of bounded variation, etc? Rich (talk) 03:27, 20 December 2020 (UTC) Another question: could we say that the real numbers under addition, using base 10 decimals is isomorphic to a uniformly bounded subgroup of G, and therefore a proper subgroup of the subgroup H defined above? Rich (talk) 03:44, 20 December 2020 (UTC)[reply]

A partial contribution. There is a result by Georg Nöbeling ("Verallgemeinerung eines Satzes von E. Specker (Generalization of a Theorem by E. Specker)". Inventiones mathematicae 6 (1968), pp. 41–55) establishing that H is free. He used the letter F for this subgroup. One publication (Rüdiger Göbel and Saharon Shelah. "How rigid are reduced products?". Journal of Pure and Applied Algebra 202, issues 1–3 (November 2005), pp. 230–258) refers to it as "the Nöbeling subgroup" of G, which seems to be the only use of this designation in the literature. I find it easier to think of the elements of G as being functions N → Z. Let π be a permutation of N and σ be any function N → {−1,+1}. Then the action a ↦ σ ○ a ○ π on G is an automorphism; since it preserves boundedness it is also an automorphism of H. I suspect that all automorphisms of G take this form; if true, this implies the invariance of H under all automorphisms of G. I see no way to connect decimal expansions to elements of G in a way that preserves algebraic structure; in general, one cannot add the digits of the decimal expansions of two numbers componentwise to get the decimal expansion of their sum. --Lambiam 10:09, 20 December 2020 (UTC)[reply]

yes,good point Rich (talk) 17:59, 20 December 2020 (UTC)[reply]

The automorphism (a0, a1, a2, a3, ...) → (a0, a0+a1, a0+a1+a2, a0+a1+a2+a3, ...) does not preserve H so H is not invariant under Aut(G). R has the divisibility property: if x is in R and n>0 is in N, the there is y so that ny=x. But for any element g≠0 of G there is some n>0 so that nh=g has no solution. So R (also Q by the same argument) cannot be isomorphic to a subgroup of G. --RDBury (talk) 13:53, 20 December 2020 (UTC)[reply]

thanksRich (talk) 17:59, 20 December 2020 (UTC)[reply]

An alternative proof that G does not have a subgroup isomorphic to R: according to Baer-Specker group, every countable subgroup is free abelian. But the rational numbers are not free abelian. So G has no subgroup isomorphic to Q (and a fortiori no subgroup isomorphic to R). --JBL (talk) 14:25, 22 December 2020 (UTC)[reply]