Is there any difference in the elements of a fixing group for a set X and the trivial subgroup of a group G? As in, if the fixing group is is there any other than the identity which can satisfy this? Could you give me an example if there is please?
Thank you
This may sound stupid, but I bet most of the members here can attest to it. When things seem stupid, it's probably because you're misunderstanding something. For example, if I was in your situation "I'd say. This is stupid, why would they define the 'fixing subgroup' of a set when it's really just " and the answer is invariably: they wouldn't. So, you know there must be something your missing. For example in shorthand notation people may leave out a lot of information but you can usually catch it (you want to otherwise you're misunderstanding something) by noticing that there doesn't seem to be a dependence on one of the variables. Like if someone gave you complicated looking formula you already know something fishy is happening since the right side has no dependence on the variable and so you have to sift through and figure out what they really mean.
Sorry for that little rant--there's just a few math learning tricks no one ever teaches you.
a group G acting on a set X is really just a factor group of G presented as a subgroup of Sym(X).
in tonio's example, X = G, and the factor group we are considering is G/Z(G) (which is isomorphic to Inn(G) < Aut(G) < Sym(G)).
in zoek's example, you have |X| = 3, and G = Sym(X) = S3 (in other words, X is just a "generic set of 3 elements" and the factor group we are considering is
G/{e}).
perhaps you can see that if we make "X" smaller, it becomes more likely that Fix(X) will get bigger.
for example, consider the square formed by the points a = (1,1), b = (1,-1), c = (-1,-1), d = (-1,1) in R^2. we can let D4 act on X = {a,b,c,d},
by considering D4 as the group of linear transformations {I, ρ, ρ^2, ρ^3, H, ρH, ρ^2H, ρ^3H}, where ρ =
[0 -1]
[ 1 0], and H =
[1 0 ]
[0 -1].
it should be plain that Fix(X) = {I}, whereas Fix({a,c}) = {I, ρH}.
D4 would also make a good example for tonio's illustration: Z(D4) = {I, ρ^2}, so letting D4 act on itself by conjugation,
we have Fix(D4) = {I, ρ^2}, which is non-trivial.
an extreme example is to let the action of G be trivial: gx = x, for all g in G (this amounts to taking the factor group G/G, which is isomorphic to the trivial group
{1_X} of Sym(X)). in this case, we always have Fix(X) = G, no matter what G is.
it's funny you should mention that. in another thread Drexel28 mentioned the theorem that if [G:H] is the smallest prime dividing |G|, then H is normal.
and of course, you use the action of G on the left cosets of H: g.aH = (ga)H to shown that the kernel of the induced homomorphism of G-->Sp lies in H.
since this kernel can't be {e} (|G| doesn't divide p! since p is the smallest prime factor of |G|), we have that 1 < |N| ≤ |H|, where N is the kernel.
if |N| < |H|, then [G:N] (which is the order of the image of G in Sp) > p, and in fact must be ≥ p^2 (because p is the smallest prime factor of |G|),
which is impossible since [G:N] divides p!
one therefore concludes N = H, and thus the action described is the one you have where Fix(G/H) = H.