A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$ A homework problem asked to find a short exact sequence of abelian groups
$$0 \rightarrow A \longrightarrow B \longrightarrow C \rightarrow 0$$ such that $B \cong A \oplus C$ although the sequence does not split. My solution to this is the sequence 
$$0 \rightarrow \mathbb{Z} \overset{i}{\longrightarrow}
\mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \overset{p}{\longrightarrow}
(\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \rightarrow 0$$
with $i(x)=(2x,0,0,\dotsc)$ and $p(x,y_1,y_2,\dotsc)=(x+2\mathbb{Z},y_1,y_2,\dotsc)$.
My new questions:


*

*Is there an example with finite/finitely generated abelian groups?

*If the answer to (1) is negative, will it help to pass to general $R$-modules for some ring $R$?

 A: I'm sorry, but my answer uses Ext groups, which may not be in the scope of your course.  
I don't know of a simple example with finitely generated abelian groups for the following reason: if $0 \rightarrow \mathbb Z \rightarrow E \rightarrow \mathbb Z / n \mathbb Z \rightarrow 0$ is an extension represented by $r+n \mathbb Z \in \operatorname{Ext}^1(\mathbb Z,\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$, then one can show that $E \cong \mathbb Z \oplus \mathbb Z/d\mathbb Z$ where $d$ is the highest common factor of $r$ and $n$.  If you assume this sequence doesn't split, then $r$ is not a multiple of $n$ and $d<n$, so the middle term will never be isomorphic to the direct sum of the of the first and third.
However, you can get a simple example using modules in the following way. Let $G=\langle g\rangle$ be an infinite cyclic group and let $\mathbb Z G$ be the associated commutative group ring.  Consider $0 \rightarrow \mathbb Z \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0$ a short exact sequence of $\mathbb Z G$-modules where the first map is inclusion into the first component and the second map is projection onto the second.  Let $g$ act on $\mathbb Z \oplus \mathbb Z$ by 
$\bigl(\begin{smallmatrix}
1&1\\ 0&1 
\end{smallmatrix} \bigr)$. Then $\operatorname{Ext}^1_{\mathbb Z G}(\mathbb Z, \mathbb Z) \cong \mathbb Z$ and this extension corresponds to $1$. If you let $g$ act by $\bigl(\begin{smallmatrix}
1&n\\ 0&1 \end{smallmatrix} \bigr)$ it corresponds to $n$. As long as $n>0$ this is a non-split short exact sequence where the direct sum of the outside terms are isomorphic to the middle one.  However, this example is far from elementary, and I apologize for that.
A: There is no counterexample with $A,B,C$ finitely generated abelian groups.  There is, more generally, no counterexample with $A,B,C$ finitely generated modules over any noetherian ring $R$.  
To see this, consider the exact sequence
$$0\rightarrow\mathrm{Hom}(C,A)\rightarrow\mathrm{Hom}(B,A)\rightarrow \mathrm{Hom}(A,A).$$
The original sequence splits if and only if this sequence is exact on the right.  If $A$, $B$ and $C$ are of finite length as modules, this follows immediately just by counting lengths.  Otherwise, it's enough to show exactness after localizing and completing at an arbitrary prime $P$, and for this it's enough to show exactness after tensoring with $R/P^n$, and for this you can assume the lengths are finite, which is the case we've already dealt with.
A: See the answer in this question:
In R-Mod Category, example for $B\cong A \oplus C \nRightarrow 0 \to A \to B \to C \to0$ splits.
It's similar to WillO's.
