For modules one has the isomorphism theorem $(A/C)/(B/C) \cong A/B$ for $C\leq B\leq A$. One way to remember it is through analogy with canceling of fractions. Another way to remember and prove it is to put all the modules in a 3×3 commutative diagram

$$

\begin{matrix}

C & \to & B & \to & B/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

C & \to & A & \to & A/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

0 & \to & A/B & \to & (A/C)/(B/C)

\end{matrix}

$$

where there are also zero arrows on the edges of the diagram, but I have omitted them for ease of typesetting. All the columns are exact, and the first two rows are exact, so the remaining row is exact giving the required isomorphism via the 3×3-lemma.