Showing a subgroup of $\mathbb{Z}\times\mathbb{Z}$ is cyclic. 
In the group $G = \mathbb{Z}\times\mathbb{Z}$, consider the subgroup $H$ generated by $(-5,1)$ and $(1,-5)$. I want to show that $G/H$ is cyclic and find the standard cyclic group it is isomorphic to.

I haven't much group theory experience, but understand that $G$ is a group. Firstly what is meant by $H$ being generated by the mentioned elements of $G$? I know it's the intersection of all subgroups that contain those two particular elements, but can it be thought of as all the multiples and linear combinations of the two?
And I am also confused about the rest of the question.
Edit: I think confusion lies with the definition of 'generated by'. I understand its the intersection of all these subgroups that contain the set of elements (or generators) but is there a more useful equivalent definition.

If $H$ is generated by $h_1$ and $h_2$ then $H=\{ah_1+bh_2\}$ where $a$ and $b$ are integers. If you think of $G$ as the set of points in the plane with integer co-ordinates then $H$ is the lattice of points with co-ordinates $(-5a+b, a-5b)$ where $a$ and $b$ are integers.
The elements of $G/H$ correspond to the co-sets of $H$ within $G$. Since the determinant of 
$\begin{pmatrix} -5 & 1 \\ 1 & -5 \end{pmatrix}$
is $24$, the area of the parallelogram bounded by $(0,0)$, $(-5,1)$ and $(1,-5)$ is $24$ so there are $24$ such co-sets.
Since $G$ is abelian, $G/H$ must also be abelian, so $G/H$ is an abelian group of order $24$. To show that $G/H$ is isomorphic to $C_{24}$ and not to some other abelian group with order $24$ (such as $C_{12} \times C_2$) we must find an element of $G/H$ that has order $24$. The co-set that contains the point $(0,1)$ is a candidate for this, since
$-5a+b=0 \Rightarrow b=5a \Rightarrow a-5b = -24a$
so if $k(0,1) \in H$ then $k$ must be a multiple of $24$, so the order of the $(0,1)$ co-set within $G/H$ is $24$.
