Torsion part of an Abelian group

Question

Let $A$ be an Abelian group such that $A=T\times K$, where $T$ is the subgroup made by the periodic elements of $A$.

If we now consider a subgroup $B$ of $A$, is it true that $B$ has the same property? In other words, does $B$ split on its torsion-subgroup?

If this is not the case, is there some counterexample?

Answer 1

I should make my comment into an answer. I believe that there exist examples of abelian groups $B$ in which the torsion subgroup $U$ has no complement In $B$. (In this post $\prod_{n=1}^\infty {\mathbb Z}/p^n{\mathbb Z}$ is mentioned as an example). Let $B$ be such a group and $K = B/U$.

Now we can embed $U$ into a divisible torsion group $T$, and this induces an embedding $B \to A$, where $A/T \cong K$. Since $T$ is divisible, the extension splits and $A = T \times K$.

Answer 2

Provided that you assumed $A$ to be finitely generated, yes it holds. What holds is that any finitely generated abelian group splits as you pointed. If $B$ is a subgroup of the fin. gen. ab. $A$, then $B$ is again a finitely generated abelian of rank less or equal to the rank of $A$, so again it splits.