Question

group theory:
A is a set, G is a finite subset of S_A
(S_A is the set of all the permutations of A)
let u be in A then G_u := {g in G s.t. g(u) = u}
in other words the set of all permutations in G which leave u fixed
I need to show G_u is a subgroup of G.

Answer 1

it seems to me, that if x is in G_u and y is in G_u then xu = u and yu = u implies x = y, so this set has only one element?

Answer 2

hmm x(u) is a function x at u so x(u) = u does not implie x = e?

Answer 3

imply*

Answer 4

To answer your first question, look at the example A = {1, 2, 3}, and let f be a function from A to A such that f(1) = 1, f(2) = 3, and f(3) = 2 (So f permutes 2 and 3, but leaves 1 fixed). Let g be the identity function on A, so g(1) = 1 , g(2) = 2, and g(3) = 3. Note they both leave 1 fixed, but they are not the same permutation. So x(u) = u and y(u) = u doesnt imply that x = y.

Answer 5

right ty

Answer 6

you need to think of u as a fixed element. Not varying. If the statement was "For all u, x(u)=u" then yes, x would be the identity. But thats not the case here.

Answer 7

ok right, its just one element that we are constructing this other set on.

Answer 8

so then x(y(u)) = x(u) = u so x*y in in g_u

Answer 9

Correct, so the set G_u is closed under the operation (function composition).

Answer 10

What else do you have to show?

Answer 11

inverse

Answer 12

closed with respect to inverses

Answer 13

if x(u) = u then u = x^-1(u) because x is 1-1 and onto?

Answer 14

Yeah, that will do the job. If x(u)=u, you apply its inverse to both sides to get:\[x^{-1}(x(u))=x^{-1}(u)\Longrightarrow id_A(u)=x^{-1}(u)\Longrightarrow u=x^{-1}(u)\](id is the identity function).

Answer 15

and of course x^-1 is in g because x is... great

Answer 16

I have another one that I have no idea what to do with, ill post if you would like ot take a look.