[tex] g=(1 2) (3 4 5 6)
Let G= [tex] {1, g, g^2, g^3}
Compute
Computer both stabilizer and orbit of 1,3,7 in G.
Note:
|stabilizer (x)||orbit (x)|=|G|
Stabilizers is the Action
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, but . Therefore, stabilises 1, as does the group identity. Therefore, the orbit of the element 1 is {1, 2} and its stabiliser is the subgroup . Note that 2.2=4=|G|.
Does that make sense?
[quote=Linnus;505474]
[tex] g=(1 2) (3 4 5 6)
Let G= [tex] {1, g, g^2, g^3}
Compute
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