group action

[hsong9]

1. The problem statement, all variables and given/known data
Let X be a G-set and let x and y denoted elements of X.
a) If x in X and b in G, show that S(bx) = bS(x)b-1
b) If S(x) and S(y) are conjugate subgroups, show that |Gx| = |Gy|


3. The attempt at a solution
Let S(x) = {a in G | ax=x}
Let S(bx) = {a in G | abx=x} => abx = x => bx = x => bxb-1 = xb-1

bS(x)b-1 = baxb-1 = bxb-1 = xb-1

Thus S(bx) = bS(x)b-1

b) Since S(x) and S(y) are conjugate subgroups, |S(x)| = |S(y)|
so by |Gx|=|G:S(x)|, |Gx|=|Gy|
correct???

[matt grime]

Let S(x) = {a in G | ax=x}
Let S(bx) = {a in G | abx=x}
=> abx = x
=> bx = x
=> bxb-1 = xb-1


No, that does not hold. That abx=x does not imply that bx=x, as that would mean that b stabilizes x which is certainly not (necessarily) true.



bS(x)b-1 = baxb-1 = bxb-1 = xb-1


Whoa, there. You've now equated a group with an element in a group.

Start again with b^-1abx = x and see what you can say.