$\displaystyle \chi = {1,2,3,4,5,6,7} $

[tex] g=(1 2) (3 4 5 6)

Let G= [tex] {1, g, g^2, g^3}

Compute $\displaystyle g^2 and g^3$

Computer both stabilizer and orbit of 1,3,7 in G.

Note:

|stabilizer (x)||orbit (x)|=|G|

Stabilizers is the Action

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- May 2nd 2010, 11:32 AMLinnusOrbits and Groups
$\displaystyle \chi = {1,2,3,4,5,6,7} $

[tex] g=(1 2) (3 4 5 6)

Let G= [tex] {1, g, g^2, g^3}

Compute $\displaystyle g^2 and g^3$

Computer both stabilizer and orbit of 1,3,7 in G.

Note:

|stabilizer (x)||orbit (x)|=|G|

Stabilizers is the Action - May 2nd 2010, 12:36 PMSwlabr
The orbit of an element in the set the group is acting upon, S, is everywhere in S it is sent to by an element of the group. So, for example, there is no way of sending 1 to anything other than 2. Similary, g sends 1 to 2.

The stabiliser of an element is the set of all group elements which "stabilise" the element; group elements which don't send the element anywhere.

So, for example, $\displaystyle 1g=2 = 1g^3$ but $\displaystyle 1g^2 = 1 = 1e$. Therefore, $\displaystyle g^2$ stabilises 1, as does the group identity. Therefore, the orbit of the element 1 is {1, 2} and its stabiliser is the subgroup $\displaystyle \{e, g^2\}$. Note that 2.2=4=|G|.

Does that make sense? - May 2nd 2010, 12:52 PMLinnus
Thanks for the help. I understand that part, but I don't know how to compute the orbit and the stabilizer of 3. Any help is appreciated!

- May 2nd 2010, 01:03 PMrubic
[quote=Linnus;505474]$\displaystyle \chi = {1,2,3,4,5,6,7} $

[tex] g=(1 2) (3 4 5 6)

Let G= [tex] {1, g, g^2, g^3}

Compute $\displaystyle g^2 and g^3$

g^2 = (1 2) (3 4 5 6)(1 2) (3 4 5 6) = (3 5)(4 6)

g^3 = (3 5)(4 6)(1 2) (3 4 5 6) = (3 6 5 4)

stab(1) = {g^2,g^3}, orb(1) = 2

stab(3) = {1}, orb(3) = 4

stab(7) = {1,g,g^2,g^3}, orb(7) = 1

i am not sure, but thats what i would do - May 2nd 2010, 01:07 PMrubic
- May 3rd 2010, 12:33 AMSwlabr