I study by myself algebra first undergraduate course. I follow the book (https://www.springer.com/us/book/9783319452845#reviews). I am trying to understand decomposability of modules (chapter 14). I try to solve the exercise 14.15. which is as follows
For $M=\mathbb{Z}^2$ and the $\mathbb{Z}$-submodule $N \subset M$ generated by the vectors (2,1) and (1,2), show that $N \simeq \mathbb{Z}^2, M/N \simeq \mathbb{Z}/(3)$, and there exists no $\mathbb{Z}$-submodule $L \subset M$ such that $M=L \oplus N$.
I had showed that the homomorphism $\phi: N \rightarrow \mathbb{Z}^2$, which send (1,0) to (2,1) and (0,1) to (1,2) is isomorphism. About $ M/N \simeq \mathbb{Z}/(3)$ I don't know how exactly I can show it formally since I don't feel I understand the proper structure of all this thing intuitively. Though I can draw the two vectors (2,1) and (1,2) on plane and count how many points of $M=\mathbb{Z}^2$ is inside the parallelogram, constructed by these vectors. And so the classes of these points can not be expressed by linear combination of (1,2) and (2,1). It seems something like quotient $M/N$. And why there exists no $\mathbb{Z}$-submodule $L \subset M$ such that $M=L \oplus N$
And how do we recognize what is the quotient of an object (group, ring, vector space, module...) when we have just an arbitrary its subset. Or we alway can construct a homomorphism which has that subset be the kernel?
I'm sorry my English is maybe not good enough, thank you for any help.
Notice that the members of $N$ are the vectors $(a,b)$ such that $a+b = 0 \mod 3$. So $N$ is the kernel of the homomorphism $M \rightarrow \mathbb{Z}/(3): (a,b) \mapsto a+b \mod 3$.
If there were a submodule $L$ such that $M=L \oplus N$ then this homomorphism would map $L$ to a proper non-trivial subgroup of $\mathbb{Z}/(3)$. But $\mathbb{Z}/(3)$ has no proper non-trivial subgroups ...