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Math Help - Fixing Groups

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    Fixing Groups

    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 \left\{  g \in G : gx=x\right\} is there any g \in G other than the identity which can satisfy this? Could you give me an example if there is please?

    Thank you
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  2. #2
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    Quote Originally Posted by Conn View Post
    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 \left\{  g \in G : gx=x\right\} is there any g \in G other than the identity which can satisfy this? Could you give me an example if there is please?

    Thank you


    Piece of cake: for any group acting on itself by conjugation, any element in the group's center fixes all the elements of
    the group. Now just choose a group with a non-trivial center and, preferably, non abelian so that the example will be cute.

    Tonio
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    Quote Originally Posted by Conn View Post
    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 \left\{  g \in G : gx=x\right\} is there any g \in G other than the identity which can satisfy this? Could you give me an example if there is please?

    Thank you
    another example:

    if G = S_3 = \left\{ 1, (12),(13), (23), (123), (132) \right\} then Fix_{S_3} (2) = \left\{ 1, (13) \right\}  .
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Conn View Post
    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 \left\{  g \in G : gx=x\right\} is there any g \in G 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 X when it's really just \{e\}" 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 \text{Ind}^G_H(\rho)(n,g)=\sum_{t\in\mathcal{T}}x_  {P_1(g,t)}\rho_{P_2(g,t)}(v_{t}) you already know something fishy is happening since the right side has no dependence on the variable n 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.
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  5. #5
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    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.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    I don't think anyone has mentioned this yet, but every normal subgroup of a group is a fixing group in a natural way.

    Let N be normal in G, and let G act on the right cosets of G/N by right multiplication. Then Nx.g=Nx if and only if g \in N.
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  7. #7
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    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.
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