(Might get some things wrong, just started)
Let the group be dihedral group $D_8$ and the set be set of $2$-subsets of $\{1,2,3,4\}$, which is $\{\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}\}$.
How would I go about getting the ${\rm orb}_{D_8}(\{1,2\})$, because I just used vertices of a square and got $\{\{4,1\}, \{3,4\}, \{2,3\}, \{1,2\}, \{4,3\}, \{2,1\}, \{1,4\}, \{3,2\}\}$, but orbits should partition the set, and orbit of the first element already has more elements than the set itself.
Does $\{1,4\}$ and $\{4,1\}$ count as the same element? Or am I doing something else wrong here?
Just a whole lot of confusion . . .
Yes.
They are the same set.
For any set $S=\{ a, b\}$, we have, by definition of a set, that $S=\{ b, a\}$.
More generally, a set $\mathcal{S}$ is considered equal to any set $\mathcal{T}$ whose elements are a permutation of the elements of $\mathcal{S}$ and there are no other and no fewer elements in $\mathcal{T}$.