subgroup is saturated iff it is a direct summand Let $A$ be a free abelian group of finite rank. Call a subgroup $B \le A$ saturated if for all $a \in A$ and positive integers $n$ such that $na \in B$, the element $a$ belongs to $B$. Call a subgroup $B \le A$ a direct summand if there is another subgroup $C \le A$ such that the natural map $B \oplus C \to A$ is an isomorphism.
Now, I postulate that a subgroup $B \le A$ is saturated if and only if it is a direct summand. What are the tools I would need to use to show this?
 A: Saturated means that $A/B$ is torsion-free. 
Since finitely generated torsion-free abelian groups are free, the exact sequence $0\to B\to A\to A/B\to 0$ is split. This shows that $B$ is a direct summand of $A$. 
The converse is obvious: if $A=B\oplus C$, and $a=b+c$ is such that $na\in B$, then $nc=0$, so $c=0$.
A: Let $A$ be a free abelian group of finite rank. Let $B \le A$ be a subgroup of $A$. notice that $A$ is a finitely generated abelian group, so $B$ is also finitely generated and torsion, and thus a finitely generated group of finite rank. Consider $A/B$. Then $A/B$ is a finitely generated group, since if $x_1, \dots, x_n$ generate $A$, then $f(x_1), \dots, f(x_n)$ generate $A/B$ where $f: A \to A/B$ is the canonical map.
Suppose $B$ is saturated. Notice that $A/B$ is torsion-free, since if $x + B$ is a torsion element with $n(x+ B) = B$, then $nx \in B$. But then $B$ saturated implies $x \in B$, that is, the only torsion element is the identity. So $A/B$ is a torsion-free finitely generated group, and thus it is free. Let $y_1 + B, \dots, y_m + B$ be a set of free generators for $A/B$, and let $C$ be the submodule of $A$ generated by $y_1, \dots, y_n$ $($notice we are implicitly choosing representatives for the cosets$)$. We claim $B \oplus C = A$. If $x \in B \cap C$, then $f(x) = B$ $($since $x\in B$, but $x \in C$ implies $$x = a_1y_1 + \dots + a_my_m$$ with $a_i \in \mathbb{Z}$$)$. Then $$B = f(x)a_1(y_1 + B) + \dots + a_m(y_m + B).$$ But since $y_1 + B, \dots, y_m + B$ is a free generating set, then $a_1 = \dots = a_m = 0$, and $x = 0$. So $B$ and $C$ intersect trivially. Suppose $a \in A$. Then $$f(a) = a_1(y_1 + B) + \dots + a_m(y_m + B),$$with $a_i \in \mathbb{Z}$. Then $$a_1y_1 + \dots + a_my_m \in C,$$ and $$a - (a_1y_1 + \dots + a_my_m) \in B$$ $($since it is in the kernel of $f)$. So $a \in B \oplus C$, and thus $A = B \oplus C$.
Suppose that $A = B \oplus C$ and $nx \in B$. Let $x = b + c$, where $b \in B$, and $c \in C$. Then $nc = nx - nb \in B$. Also $nc \in C$. So $nc \in B \cap C = \{0\}$. But $A$ is torsion-free, so $c = 0$, and $x = b \in B$, that is, $B$ is saturated.
A: If you think of an Abelian group as a $\mathbb{Z}$-module, then what you call a saturated subgroup is exactly what is usually referred to as a pure submodule. The structure theory of finitely generated modules over a principal ideal domain comes into play here. You don't need the full strength of it for this question, but a  sharper version of the general structure theorem (which is stated in many algebra texts) is that if $R$ is a prinicipal ideal domain,  and $N$ is a submodule of a finitely generated $R$-module $M,$ then $M$ has an $R$-basis $\{ m_{1},m_{2}, \ldots m_{r}, m_{r+1},\ldots m_{s} \}$ such that for some set of elements $d_{1}, d_{2}, \ldots d_{r} \in R$, $\{d_{1}m_{1}, d_{2}m_{2},\ldots d_{r}m_{r} \}$ is an $R$-basis for $N.$ If $N$ is a pure submodule of $M,$  then each $d_{i}$ must be a unit of $R,$ and then $N$ is a direct summand of $M.$  Applying this with $R = \mathbb{Z}$, $A$ in the role of $M$, $B$ in the role of $N$ gives what you want ( the inclusion "$B$ direct summand implies $B$ saturated" being obvious).
