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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.

MyWorld
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    It's simple: $G$ is not a group. For instance $(1234)^2=(13)(24)$ is not in $G$. – Captain Lama Apr 29 '19 at 16:47
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    I think you made a confusion with the double-transposition group. – Captain Lama Apr 29 '19 at 16:48
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    I think they are using a linear ordering notation for the permutations rather than a cycle notation (i.e. they are giving the images of $1,2,3,4$ in that order). For example $(1234)$ is the identity, $(2134)$ is the transposition $(1,2)$, and $(2143)$ would be, in cycle notation $(1,2)(3,4)$. Then what they say about orbits is right, and moreover, in cycle notation, the four given permutations don't form a group. – Ned Apr 29 '19 at 16:52
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    @CaptainLama There are various notational conventions for permutations. Here certainly the one-line notation is used. See https://en.wikipedia.org/wiki/Permutation. – Paul Frost Apr 29 '19 at 16:56

1 Answers1

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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.

Santana Afton
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