$0 \rightarrow Q \rightarrow M \rightarrow P \rightarrow 0$ is isomorphic to some $0 \rightarrow N \rightarrow M \rightarrow M/N \longrightarrow 0$ Let $A$ be a ring and $M, Q, P$ be $R$-modules. Let $N$ be a submodule of $M$. Then any short exact sequence of the form $0 \longrightarrow Q \longrightarrow M \longrightarrow P \longrightarrow 0$ is isomorphic to the exact sequence an exact sequence  $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0$. Why is that the case? I imagine we somehow consider the diagram 
\begin{array}{ccccccccc}
  0 &\longrightarrow & Q & \longrightarrow & M & \longrightarrow & P & \longrightarrow & 0\\
  & & \downarrow & & \downarrow & & \downarrow & & \\
  0 &\longrightarrow & N & \longrightarrow & M & \longrightarrow & M/N &  \longrightarrow & 0
\end{array}
and want to construct some isomorphism between $P$ and $M/N$, and between $Q$ and $N$, but I'm having a hard time seeing how we could do so.
 A: What it means for $0 \longrightarrow Q \longrightarrow M \longrightarrow P \longrightarrow 0$ to be a short exact sequence is that

(1) $Q \rightarrow M$ is injective.
(2) $\text{im}(Q \rightarrow M) = \ker(M \rightarrow P)$
(3) $M \rightarrow P$ is surjective

Informally, we get the sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0$ by conceiving of $Q$ as a submodule of $M$ and $P$ as a projection of $M$.
To be precise, we're just going to apply the first isomorphism theorem for modules twice.  As you suspected, this produces canonical isomorphisms $Q \cong N$ and $P \cong M/N$.  After establishing the isomorphisms, it will just remain to check that the sequences themselves are isomorphic, which amounts to checking that the diagram you gave above is commutative.
Let's set $N = \text{im}(Q \rightarrow M)$.  Then the first isomorphism theorem guarantees $Q / \ker(Q \rightarrow M) \cong \text{im}(Q \rightarrow M) = N$, and by $(1)$, the kernel is trivial so that $Q \cong N$.
Similarly the first isomorphism theorem guarantees $M / \ker(M \rightarrow P) \cong \text{im}(M \rightarrow P)$.  Applying both $(3)$ $\big(\text{im}(M \rightarrow P) =P \big)$, and $(2)$ $\big(\ker(M \rightarrow P) = N\big)$ we have $M / N \cong P$.
A: Hint: The diagram you are looking for is
$\require{AMScd}$
\begin{CD}
    0 @>>> Q @>i>> M @>p>> P @>>> 0\\
    @.       @ViVV @V\operatorname{id}_MVV @V\varphi VV \\
    0 @>>> \operatorname{im} i @>>> M @>>> M/\operatorname{im} i @>>> 0
\end{CD}
Where $\varphi$ is the unique map such that $(\varphi\circ p)(m) = m + \operatorname{im}i$. To prove that such $\varphi$ exists (and is unique), use that $p$ is epimorphism and $\operatorname{im} i = \ker p$.
