Here is the question I am trying to solve (I know it is answered here $A$ be a nonzero finite abelian group then $A$ is not a projective or injective $\Bbb Z$ module. but the answer is not very clear to me, so I am trying to write every detail in my solution below and ask for clarification of steps that I do not know how to prove):
Let $A$ be a nonzero finite abelian group.
$(a)$ Prove that $A$ is not a projective $\mathbb Z$-module.
My thoughts:
Since $A$ is a nonzero finite abelian group, then by the fundamental theorem of finitely generated abelian groups, we know that $A \cong \mathbb Z^r \times \mathbb Z_{n_1} \times \mathbb Z_{n_2} \times \dots \times \mathbb Z_{n_s}$ with $r \geq 0$ and $n_j \geq 2$ for all $j$ and $n_{i+1}|n_i$ for $1 \leq i \leq s-1.$
But then I want to use the following definition of a projective module to negate that $A$ is projective:
If $P$ is a quotient of the $R$-module $M$ then $P$ is isomorphic to a direct summand of $M,$ i.e., every short exact sequence $$ 0 \to L \to M \to P \to 0 $$ splits.
Where $L$ is an $R$-module.
I am wondering why there is a surjection $\pi: \mathbb Z^r \to A$ and why if $\mathbb Z^r$ does not have torsion elements this map cannot split? What exactly the exact sequence that we are using?
Can explain this in detail to me please?
First, note that since $A$ is finite $r=0$.
Second, Consider the map $f:\mathbb{Z}\rightarrow A:1\rightarrow (1,1,\ldots,1)$. This map is surjective so you can build an exact sequence out of it, but $f$ does not split and you reach a contradiction.