Quotient of pure subgroups I follow Rotman's textbook "Introduction to homological algebra".
This is the exercise 3.45 in the book.

Let $G$ be an abelian group, and let $S\subseteq G$ be a pure subgroup. If $S\subseteq H\subseteq G$, prove that $H$ is a pure subgroup of $G$ if and only if $H/S$ is a pure subgroup of $G/S$.

I tried to solve it but I couldn't. If you know, please explain about that.
 A: To begin with let me try to provide you with a very important remark:

Proposition: Given abelian group $G$ which is flat as a $\mathbb{Z}$-module and subgroup $H \leqslant G$, then $H$ is a pure subgroup of $G$ if and only if $G/H$ is a flat $\mathbb{Z}$-module.

Since for $K \leqslant H \leqslant G$ you canonically have $G/H \approx (G/K)/(H/K)$, this does offer some insight on the equivalent statements in the exercise.
It is possible to supply a domestic answer, based on the definition and a few elementary facts about mappings: recall that in general any map $f: A \to B$ between arbitrary sets will induce a canonical equivalence $R$ on $A$, defined by $xRy \Leftrightarrow f(x)=f(y)$; we will denote this equivalence by $\mathrm{Eq}(f)$. 
Also recall that given subsets $X, Y \subseteq A$ such that say $X$ be saturated with respect to $\mathrm{Eq}(f)$ (this means that along with any element $x$, $X$ contains the entire class of $x$ with respect to $\mathrm{Eq}(f)$), then we have 
$$f(X \cap Y)=f(X) \cap f(Y) \tag{satint}$$
Consider now an abelian group $G$ together with subgroups $K \leqslant H \leqslant G$. Assuming that $H$ is pure in $G$ means by definition that for any $n \in \mathbb{Z}$ we have 
$$nG \cap H=nH \tag{1}$$ 
Introduce the canonical surjection $\sigma: G \to G/K$ and note that its canonical equivalence is none other than the congruence modulo $K$, with respect to which $H$ is saturated (since $H \supseteq K$); therefore, by applying $\sigma$ to relation (1) and bearing in mind that relation (satint) holds here, one obtains 
$$n(G/K) \cap (H/K)=n(H/K) \tag{2}$$
which is none other than the statement of the purity of $H/K$ as a subgroup of $G/K$. 
Conversely, if in addition to the purity of $K$ in $G$ one assumes this latter purity involving the quotients, expressed by relation (2), the same property (satint) tells one that this relation equivalently becomes
$$\sigma(nG \cap H)=\sigma(nH) \tag{3}$$
and by taking inverse images through $\sigma$ one obtains
$$(nG \cap H)+K=nH+K \tag{4}$$
Taking in relation (4) the intersection with subgroup $nG$ and bearing in mind the property of modularity (that the lattice of subgroups of $G$ possesses), one has
$$(nG \cap H)+(nG \cap K)=nH+(nG \cap K)$$
which by the purity of $K$ in $G$ is reduced to
$$(nG \cap H)+nK=nH+nK \tag{5}$$
Since clearly $nK \leqslant nG \cap H, nH$, relation (5) leads right away to the conclusion that $H$ is pure in $G$.
