Let $G = \mathbb Z \times \mathbb Z = \mathbb Z^2 $ and let $H$ be the subgroup generated by $(1,3)$ and $(2,1)$, i.e., $$H = \{ m(1,3) + n(2,1) \, : \, m,n \in \mathbb Z\}.$$ This exercise will help you understand $G / H$ concretely.
1) Now let $(x,y) \in \mathbb Z^2$. Because $(1,3)$ and $(2,1)$ span $R^2$ there are real numbers $s$ and $t$ such that $(x,y) = s (1,3) + t (2,1)$. Let $(a,b) = (s - \lfloor s\rfloor) (1,3) + (t - \lfloor t \rfloor) (2,1)$. Prove that $(a,b) \in \Gamma$ and $(x,y) + H = (a,b) + H$.
2) Argue that if $(a,b)$ and $(c,d)$ are distinct elements in $\Gamma$ then the cosets $(a,b) + H$ and $(c,d) + H$ are distinct.Conclude that $G / H = \{ (a,b) + H \, : \, (a,b) \in \Gamma \}$.
3) Why is $G/H$ a cyclic group? Can you find a generator?
Solution:
1) We know that $(a,b) = (s - \lfloor s\rfloor) (1,3) + (t - \lfloor t \rfloor) (2,1)$. Since $0 \le (s - \lfloor s\rfloor) < 1$ and $0 \le (t - \lfloor t \rfloor) < 1$ then we can let $(s - \lfloor s\rfloor) = \lambda$ and $ (t - \lfloor t \rfloor) = \mu$. Then $(a,b)\in P$.\ Since $(x,y) = s (1,3) + t (2,1)$ then $x = s + 2t$ and $y = 3s + t$.So we now have,\ $(a,b) = (s - \lfloor s\rfloor) (1,3) + (t - \lfloor t \rfloor) (2,1)$\ \indent =$(s - \lfloor s\rfloor + 2t - 2\lfloor t\rfloor, 3s - 3\lfloor s\rfloor + t + \lfloor t\rfloor)$\ \indent =$(s + 2t - \lfloor s\rfloor - 2\lfloor t\rfloor, 3s + t - 3\lfloor s\rfloor - \lfloor t\rfloor)$\ Let $j = - \lfloor s\rfloor - 2\lfloor t\rfloor \in \mathbb Z$ and $k = - 3\lfloor s\rfloor - \lfloor t\rfloor \in \mathbb Z$ since $ \lfloor s\rfloor, 2\lfloor t\rfloor, 3\lfloor s\rfloor, \lfloor t\rfloor \in \mathbb Z$. And since we know that $x = s + 2t$ and $y = 3s + t$ then\ $(a,b) = (x + j, y + k) \in \mathbb Z^2$. Thus $(a,b) \in P \cap \mathbb Z^2 = \Gamma$.\
2) We now must prove that $(x,y) + H = (a,b) + H$. By coset theorem, if $(x,y)-(a,b) \in H$, then $(x,y) + H = (a,b) + H$ is true. So,\ $(x,y)-(a,b) = s(1,3) + t(2,1) - [(s - \lfloor s\rfloor) (1,3) + (t - \lfloor t \rfloor) (2,1)]$\ = $(s + 2t, 3s + t) - [(s + 2t - \lfloor s\rfloor - 2\lfloor t\rfloor, 3s + t - 3\lfloor s\rfloor - \lfloor t\rfloor)] $\ = $(\lfloor s\rfloor + 2\lfloor t\rfloor, 3\lfloor s\rfloor - \lfloor t\rfloor) \in H$. Therefore, $(x,y) + H = (a,b) + H$.\
We know that $(a,b)$ and $(c,d)$ are distinct in $\Gamma$. We want to show that $(a,b)+H \ne (c,d)+H$. We need to show that $(a,b)-(c,d)$ is not in $H$. $\Gamma = \{(0,0), (1,1), (1,2), (2,2), (2,3)\}$. Since $\Gamma$ is not in $H$ then the difference of any two distinct pairs in $\Gamma$ are not in $H$. Thus, $(a,b)-(c,d)$ is not in $H$ and $(a,b) + H$ and $(c,d) + H$ are distinct. Since H is a subgroup of G, then $G / H = \{ (a,b) + H \, : \, (a,b) \in \Gamma \}$.
3) i am unsure on how to answer this.
(I dont think i am missing anything on 1 and 2 but can you please check and can anyone help me with 3 please?)
Your argument for 2 isn't quite complete. For example, the statement:
by which I presume you mean:
isn't fully justified. For example, what if it were the case that $\Gamma=\{(1,1),(2,4)\}$? Since $\Gamma$ is finite, you could just calculate all differences, and verify none are in $H$. A better way would be to note that $\Gamma\subset P$, so we can write
$$(a,b)=\lambda_1(1,3)+\mu_1(2,1),\quad(c,d)=\lambda_2(1,3)+\mu_2(2,1)$$
where $0\leq\lambda_1,\lambda_2,\mu_1,\mu_2<1$. So the difference of $(a,b)$ and $(c,d)$ is
$$(\lambda_1-\lambda_2)(1,3)+(\mu_1-\mu_2)(2,1)$$
Since $-1<(\lambda_1-\lambda_2),(\mu_1-\mu_2)<1$, and $(a,b)-(c,d)\neq0$, this difference cannot lie in $H$.
Now you have shown that all elements of $G$ are in a coset of the form $(a,b)+H$, where $(a,b)\in\Gamma$. You have also shown that distinct elements of $\Gamma$ produce different cosets. This tells you that the cosets bijectively correspond to elements of $\Gamma$, from which it follows that
$$G/H=\{(a,b)+H\mid (a,b)\in\Gamma\}$$
and moreover all the listed elements in that set are distinct.
Question 3) isn't too bad, given what you have already found out. You listed the 5 elements of $\Gamma$, and due to the bijective correspondence I previously mentioned this is equal to the number of cosets of $H$ in $G$, i.e.
$$|G/H|=5.$$
As groups of prime order are cyclic, then so is this one. Moreover all non identity elements are generators.
To perform arithmetic in $G/H$, you essentially are working with elements of $\Gamma$. All operations are identical to the operations in $G$, except if you leave the set $\Gamma$, you subtract an element of $H$ to get back in (this will give the same coset). So we could use this to show that $(1,1)+H$ generates $G/H$:
So you can see that $(1,1)$ generates the group explicitly.