I think this is a nice problem:
Problem. Explicitly describe the (ring of) endomorphisms of a finitely generated abelian group $\displaystyle G.$
The answer should be in a matrix form.
Well, almost by definition the set of $\displaystyle m\times m$ matrices with entries from $\displaystyle \mathbb{Z}_{n^a}$ form the endomorphism ring of $\displaystyle \mathbb{Z}_{n^a}^m$ (with $\displaystyle n\in\mathbb{N}\cup\{\infty\}$, $\displaystyle \mathbb{Z}_{\infty}:=\mathbb{Z}$). So the question is "how do we pin these together?" Basically, you pin them together by assuming you have your fgab in a "collected" form, with different copies of isomorphic groups beside each other,
$\displaystyle G\cong \mathbb{Z}_{n_1^{a_1}}^{m_1}\times\mathbb{Z}_{n_2^ {a_2}}^{m_2}\times\ldots\times\mathbb{Z}_{n_p^{a_p }}^{m_p}$ where $\displaystyle \mathbb{Z}_{n_i^{a_i}}$ appears nowhere else in the decomposition of your group.
Then, your ring of endomorphisms is the ring of $\displaystyle m_1+m_2+\ldots+m_p$ matrices, where the top $\displaystyle m_1$ elements on the diagonal, and the $\displaystyle m_1\times m_1$ square around them, have entries from $\displaystyle \mathbb{Z}_{n_1^{a_1}}$, the $\displaystyle (m_1+1)^{th}$ to the $\displaystyle m_1+m_2$ elements on the diagonal, and the $\displaystyle m_2\times m_2$ square around them have entries from $\displaystyle \mathbb{Z}_{n_2^{a_2}}$, etc. Everywhere else is zero.
Clearly all such matrices form a ring, and clearly they are endomorphisms of the abelian group. It is then sufficient to prove that every endomorphism appears in this way, but that is routine...and would take too long to type (I'm lazy).
yes, we first use the fundamental theorem for finitely generated abelian groups and then, for abelian groups $\displaystyle G_i,$ we apply the ring isomorphism $\displaystyle \text{End} \left (\bigoplus_{i=1}^n G_i \right ) \cong R,$ where $\displaystyle R$ is the matrix ring whose $\displaystyle (i,j)$-entry is $\displaystyle \text{Hom}(G_j,G_i).$ so we just need to find $\displaystyle \text{Hom}(G_j,G_i),$ where $\displaystyle G_i$ and $\displaystyle G_j$ are cyclic groups (finite or infinite). that's the idea but the full solution is quite long.