Group orbit: confusion

group-theorypermutations

Definition:

If $x \in X$, the set $x^G = \{ x^g \in X: g \in G\}$ is called the orbit of $x$ under the action of $G$.

I wanted to look up some example and stumbled upon this one:

For the permutation group $G = \{(1234),(2134),(1243),(2143)\}$ the
orbit of $1$ and $2$ is $\{1,2\}$ and the orbit of $3$ and $4$ is
$\{3,4\}$.

(Link: http://mathworld.wolfram.com/GroupOrbit.html).

By looking at the provided definition, I would have guessed that the orbit of $1$ for example is $\{2,3,4\}$. My reasoning: take $1 \in X$, and take the first permutation in $G$. This one maps $1$ t0 $2$. The second permutation in $G$ maps $1$ to $3$, and so on. Meaning that the set of possible images of $x$ under the action of $G$ equals $\{2,3,4\}$.

Why is my reasoning incorrect?

Thanks.

Best Answer

The notation Wolfram is using isn’t cycle notation; see Paul Frost’s comment. The group $G$ they give is, using cycle notation, the group

$$\{e, (1~2), (3~4), (1~2)(3~4)\}.$$

Hopefully this makes it a bit more clear why the orbits are what they are.