Ext and extensions There are two abelian groups up to isomorphism of order $p^2$, where $p$ is a prime.
But Ext$(\mathbb{Z}/p,\mathbb{Z}/p)$ is cyclic of order $p$.
I can embed $\mathbb{Z}/p$ into $\mathbb{Z}/p^2$ in $p-1$ distinct ways, inducing nonhomologous cocycles corresponding to the same abstract group.
How do I rescue my intuition of "Ext = extensions" in the light of this fact?
 A: To have an isomorphism of extensions in $\operatorname{Ext}(A,B)$, you need a commutative diagram like this:
$$\require{AMScd}\begin{CD}
0 @>>> B @>>> E @>>> A @>>> 0 \\
@. @| @VfVV @| \\
0 @>>> B @>>> E' @>>> A @>>> 0
\end{CD}$$
Then the five lemma implies that $f$ is an isomorphism of abelian groups. So if two extensions are isomorphic, the underlying abelian groups are isomorphic. But the converse need not be true, as you noticed: in $\operatorname{Ext}(\mathbb{Z}/3, \mathbb{Z}/3)$, there are two nonisomorphic extensions with underlying abelian group $\mathbb{Z}/9$. They are of the form
$$0 \to \mathbb{Z}/3 \xrightarrow{\varphi_i} \mathbb{Z}/9 \xrightarrow{p} \mathbb{Z}/3 \to 0$$
where $p$ is the reduction mod $3$, and $\varphi_i(1) = i$ for $i = 3$ and $i = 6$. Then it's impossible to find a self-map $f : \mathbb{Z}/9 \to \mathbb{Z}/9$ such that at the same time $p \circ f = p$ and $f \circ \varphi_3 = \varphi_6$ (or the reverse), something that I encourage you to check. Thus the extensions are not isomorphic, even though the underlying abelian groups are.
