Abelian group generated by two elements subject to two relations isomorphic to a quotient group We were given the following example in lectures:
Consider the abelian group $G$ generated by $a$ and $b$ such that the relations
$2a + b = 0 $
$-a + 2b = 0$
hold.
Then $G \cong \frac{\mathbb{Z}^2}{N}$ where $N$ is generated by $(2,1)$ and $(-1,2)$.
Question: I can't seem to understand why this is the case.
Attempt at a solution: I have tried to write an explicit isomorphism e.g.
$\phi : G \rightarrow \frac{\mathbb{Z}^2}{N}$
$a \mapsto (2,1)$, $b\mapsto (-1,2)$ - does this even work?
Alternatively this might just be a case of applying the isomorphism theorem, but I'm not sure how to surject $G$ into $\mathbb{Z}^2$....
Any help would be appreciated!
 A: The free Abelian group $F$ generated by two elements $a,b$ is isomorphic to $\Bbb Z^2$ by sending $a$ to $(1,0)$ and $b$ to $(0,1)$.
Then the relation $2a+b\ \sim\ 0$ turns into $(2,1)\sim(0,0)$, so in order to obtain it as equality, we consider the quotient $F/\langle 2a+b\rangle$, or, via the isomorphism, $\Bbb Z^2/\langle (2,1)\rangle$.
And similarly for the other relation.
The first isomorphism theorem is a good attempt, but do it the other way around: look for a surjective map $\Bbb Z^2\to G$ whose kernel is just $\langle (2,1),\,(-1,2)\rangle$.
A: Write an explicit isomorphism in the other direction. Start with
$$
\psi : \mathbb Z^2 \to G,
\qquad
\psi(1,0)=a,
\qquad
\psi(0,1)=b
$$
or $\psi(x,y)=xa+yb$.
Then $\psi$ is a surjective homomorphism with kernel $N$. This latter part is the one that needs careful checking.
Here is an ad hoc solution.

*

*$b=-2a$ implies that $G = \langle a,b \rangle = \langle a \rangle$.


*$a=-2b$ implies $5a=0$ and so $a=0$ (and $G$ is trivial) or $a$ has order $5$. Let's assume the latter.


*$\psi(x,y)=xa+yb=(x-2y)a$ and so $\psi(x,y)=0$ iff $x-2y \equiv 0 \bmod 5$. Write $x-2y=5k$.


*We want to write $(x,y)=u(2,1)+v(-1,2)$. The solution is
$$
u = (2 x + y)/5 = 2k + y ,\quad v = (2 y - x)/5  = -k
$$
