For the permutation group $G=\{(1234),(2134),(1243),(2143)\}$ on the set $ X=\{1,2,3,4\}$. The orbits of $1,2$ are$ \{1,2 \}$. How the calculation is done here?
I know that orbit of an element $x \in G$ is defined by $ G(x)=\{gx \in X| g \in G\}$, where $X=\{1,2,3,4\}$.
How do I multiply a permutation of $G$ by $1$ or $2$?
Permutations in $G$ are functions $X \to X$. Therefore the group action of $G$ on $X$ is naturally identified with function evaluation.
For instance, the element $(2134) \in G$ is the function $X \to X$ defined by $$(2134)(1) = 2, \qquad (2134)(2) = 1 \\ (2134)(3) = 3, \qquad (2134)(4) = 4$$
To evaluate the orbit of $1 \in X$ under the group action of $G$, just evaluate each of the permutations at $1$. You'll find the result is either $1$ or $2,$ and since the set of orbits of a group action forms an equivalence class, the orbit of both $1$ and $2$ is $\{1,2\}$
Note on notation: This is not cycle notation -- it cannot be. The question and the referenced article tell us that $G$ is a group, and if we were to force cycle notation on them, then $G$ is definitely not a group. Rather, the notation $(a_1a_2a_3\ldots a_n)$ is an inline shorthand for the permutation of $\{1,2,3,\ldots,n\}$ mapping $i \mapsto a_i$.