Show that the normal subgroup is cyclic Let $G=\mathbb Z\times\mathbb Z$. Consider $H\leq G$ generated by $(-5,1)$ and $(1,-5)$. Show that $\frac{G}H$ is cyclic.
This is what I have so far but I'm not sure if I'm right either.
Let $a,b,\in\mathbb Z$. So $a(-5,1)+b(1,-5)=(b-5a,a-5b).$ 
Then $H=\{(b-5a,a-5b)\equiv(b,a) \mod 5\;|a,b\in\mathbb Z\}$
So $\frac{G}H= \{[(b,a)] \mod 5\}$.
The way I see it, this is generated by $[(0,1)]$ and $[(1,0)]$. Where have I made my mistake?
 A: You can use the structure theorem for such questions. The matrix
$$
\left(\begin{matrix}
-5 & 1 \\
1 & -5
\end{matrix}\right)
$$
has Smith normal form
$$
\left(\begin{matrix}
1 & 0 \\
0 & 24
\end{matrix}\right)
$$
So $G/H\cong(\mathbb{Z}/1\mathbb{Z})\times(\mathbb{Z}/24\mathbb{Z})=\mathbb{Z}/24\mathbb{Z}$, which is cyclic.
A: Well, you have grabbed the problematic here, imo: since
$$(1,0)+H\neq H\neq (0,1)+H\;,\;\;G/H\;\;\text{is cyclic}\;\iff G/H=\langle\;(1,0)+H\;\rangle$$
and then we should get that
$$\exists\,m\in\Bbb Z\;\;s.t.\;\;(0,1)+H=m\left((1,0)+H\right)=(m,0)+H\iff$$
$$(m,-1)\in H\iff (m,-1)=(b-5a\,,\,a-5b)\;,\;\;a,b\in\Bbb Z$$
So we need to solve the diophantine linear system
$$\begin{cases}&-5a+b&=&m\\&a-5b&=&-1\end{cases}\;\;\;\implies $$
$$-24b=m-5\implies\;(i)\; b=\frac{5-m}{24}\;,\;\;\text{and also}\;(ii)\;a=\frac{b-m}{5}=\frac{\frac{5-m}{24}-m}{5}=\frac{1-5m}{24}$$
Since $\,a,b\in\Bbb Z\,$ , we get that
$$(i)\implies m=5\pmod{24}\;\;,\;\;\;(ii)\implies 5m=1\pmod {24}\implies$$
And yes: there is a solution: $\,m=5\implies (5,0)+H=(0,1)+H\,$ , since in fact
$$(5,-1)=(0-5\cdot(-1)\,,\,-1-5\cdot 0)\;\;!$$
