# Help understanding orbits - specific example with permutations

I am trying to understand the definition of an orbit and came across another post which has a nice, concise definition:

The orbit of xx is "everything that can be reached from xx by an action of something in GG."

The stabilizer of xx is "the set of all elements of GG which don't move xx when they act on xx".

Those already seem pretty intuitive... what else can be said?

I guess you might want to look at orbits and stabilizers for particular >actions. For example, if a group is acting on itself by conjugation, then >the orbit of an element is that element's conjugacy class. One element stabilizes another in this action exactly when they commute.

However, I am confused as to how this relates to a definition and an example found here:

When a group G acts on a set X (this process is called a group action), it permutes the elements of X. Any particular element X moves around in a fixed path which is called its orbit. In the notation of set theory, the group orbit of a group element x can be defined as G(x)={gx in X:g in G}, where g runs over all elements of the group G.

For example, for the permutation group G = {(1234),(2134),(1243),(2143)}, the orbits of 1 and 2 are {1,2} and the orbits of 3 and 4 are {3,4}.

What I don't understand is why the orbits of 1 and 2 are {1,2} and why the orbits of 3 and 4 are {3,4}. Why I don't understand this is the second definition states (emphasis added):

In the notation of set theory, the group orbit of a group element x can be defined as G(x)={gx in X:g in G}, where g runs over all elements of the group G.

and as the first definition states:

The orbit of xx is "everything that can be reached from xx by an action of something in GG."

The statement "where g runs over all elements of the group G" makes me think (probably improperly) that we would obtain the orbit of element x for permutation group G by applying all elements of G to x. Concretely, by this definition I would think that given an element x, where x = 1 and a permutation group G where G = {(1234), (2134), (1243), (2143)} you would obtain the orbit by doing the following:

(1234)(1) -> 2

(2134)(1) -> 3

(1243)(1) -> 2

(2143)(1) -> 4

Therefore, the orbit G(1) = {2,3,4}. I am thinking this because, as the first definition states, I can reach the elements 2, 3, and 4 from 1 by an action of a member of group G.

If someone could correct my thinking on this and explain why, in this example, the orbits of 1 and 2 should be {1, 2} and the orbits of 3 and 4 should be {3, 4} it would be greatly appreciated.

Also, if you are feeling particular generous, I am trying to solve 3 sets of problems with regards to orbits and permutations found in question 34(c) here. If there is anything additional information/explanation required to solve these I would greatly appreciate it if I could see an example worked out. One of the three questions states:

Compute the orbits of each of the following elements in $S_5: α=(1254)$.

## 1 Answer

The confusion is in the notation. You may be expecting cycle notation, where $(1234)$ means $1$ permutes to $2$, etc. In fact, the notation wolfram is using means that each number permutes to the number in its place. Thus $(1234)$ is the identity permutation.

We could rewrite in cycle notation the group $G_1=\{e,(12),(34),(12)(34)\}$ where $e$ is the identity. Then hopefully it is more clear that the orbit of $1$ is $\{1,2\}$.

On that note, you should know that an element is always in its own orbit since each group contains an identity. Thus you should be suspicious that, as you mentioned, it looks like the orbit of $1$ doesn't contain $1$.

• That clarifies A LOT. Thank you. So, basically my understanding was right but the notation was confused? And, for further clarification, given my example $α=(1254)$, I would compute all $A = a^n = \{e, (1254), (15)(24), (1452) \}$ and apply the operation I described above? $A(1) = \{1, 2, 5, 4\}$ in this instance? Is this correct? – user1105224 Nov 3 '15 at 0:28
• Not sure I understand the example, but given the group $\{e,(1254),(15)(24),(1452)\}$ the orbit of one would be $1,2,4,5$ since one can move to any element. The orbit is maybe most intuitively "the list of elements to which X can possibly permits to" – Elliot G Nov 3 '15 at 0:36
• That's fine. I've run into an occasional oddly worded question in this book, and I think that's all this may be then. At least now I better understand the definition of an orbit and how to compute them. – user1105224 Nov 3 '15 at 1:08