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.
 A: As you seem to have perceived, this is as much a linear algebra question as a group theory question, although you have to be careful and do your linear algebra over $\mathbb Z$ instead of the usual $\mathbb R$. That means you can only use integers where you are used to using arbitrary real numbers.
The group $H$ is generated by two integer vectors $\vec v = (-5,1)$ and $\vec w = (1,-5)$. Since this is an abelian group, then yes, you can say that $H$ is the group of all integer linear combinations of $\vec v$ and $\vec w$. 
Now let's put those two vectors into the rows of a matrix:
$$M = \begin{pmatrix} -5 & 1 \\ 1 & -5 \end{pmatrix}
$$
It follows that the row space of $M$ over $\mathbb Z$ is $H$, i.e. the set of all integer linear combinations of the rows of $M$ is $H$.
Now use your linear algebra skills to simplify the matrix $M$ by doing row operations which do not affect the row space over $\mathbb Z$. For example, add $5$ times row 2 to row 1 to get
$$\begin{pmatrix} 0 & -24 \\ 1 & -5 \end{pmatrix}
$$ 
then switch rows 1 and 2 to get
$$\begin{pmatrix} 1 & -5 \\ 0 & -24  \end{pmatrix}
$$
and then multiply row $2$ by $-1$ to get
$$\begin{pmatrix} 1 & -5 \\ 0 & 24  \end{pmatrix}
$$
You can also do column operations over $\mathbb Z$, which have the effect of changing the given basis for $G$, but of course that does not affect the isomorphism type of the quotient group $G/H$. So, adding $5$ times column $1$ to column $2$ you get
$$\begin{pmatrix} 1 & 0 \\ 0 & -24  \end{pmatrix}
$$
So now we know that
$$G / H \approx (\mathbb Z \oplus \mathbb Z) / (\mathbb Z \oplus 24\mathbb Z) \approx (\mathbb Z / 1 \mathbb Z) \oplus (\mathbb Z / 24\mathbb Z) \approx \mathbb Z / 24\mathbb Z
$$
so the quotient is isomorphic to the cyclic group of order $24$.
A: 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 $24$.
