Divisible abelian subgroup of abelian group contains no proper nontrivial verbal subgroup

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Suppose G is an abelian group, and H is a subgroup of G that is a Divisible abelian group (?). Then, if K is a subgroup of H that is a verbal subgroup of G, then either K is trivial or K = H.

Facts used

  1. verbal subgroup equals power subgroup in abelian group


Given: An abelian group G, a subgroup H of G that is divisible, and a subgroup K of H that is verbal in G.

To prove: K = H or K is trivial.

Proof: By fact (1), there exists an integer n such that K is the set of n^{th} powers in G. If n = 0, K is trivial. If, on the other hand, n is nonzero, then every element of H is a n^{th} power, so H \le K. Since K \le H by assumption, we get H = K.