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Now, the question asks me what the subgroups of order $4$ are of this relation and then to give them as sets and identify the group of order $4$ that each of the subgroups is isomorphic to. How do I go about doing that? Here is what I got.

I know this relation has $8$ elements: $(0,0),\ (0,1),\ (0,2),\ (0,2),\ (0,3),\ (1,0),\ (1,1),\ (1,2),\ (1,3),$

I'm completely lost from this point. How do I find a subgroup of a relation like this?

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It is known that the only groups of order $4$ are isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \times \mathbb{Z}_2$. Let $S$ be your subgroup.

1) If $S$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, then all its elements have order $2$. But the elements of order $2$ of $\mathbb{Z}_2 \times \mathbb{Z}_4$ are $\{(0,0), (0,2), (1,0), (1,2)\}$ which are seen to form a subgroup of order $4$.

2) If $S$ is isomorphic to $\mathbb{Z}_4$, then it must be cyclic, that is it must contain at least an element of order $4$. But the elements of order $4$ of $\mathbb{Z}_2 \times \mathbb{Z}_4$ are $\{(0,1), (0,3), (1,1), (1,3)\}$. The first $2$ generate, each, the subgroup $\{(0,0), (0,1), (0,2), (0,3)\}$; the latter $2$ generate, each, the subgroup $\{(0,0), (0,2), (1,1), (1,3)\}$.

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