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Math Help - Centre of a Group and Conjugacy Classes

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    Super Member Bernhard's Avatar
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    Centre of a Group and Conjugacy Classes

    Dummit and Foote Section 4.3 Groups Acting on Themselves by Conjugation - The Class Equation - Exercise 5 reads:

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    If the centre of G is of index n, prove that every conjugacy class has at most n elements

    ================================================== =====

    I am having trouble getting started on this problem. Can anyone help?

    Peter
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    Re: Centre of a Group and Conjugacy Classes

    do you know the class equation?

    EDIT: maybe that's not so helpful, here.

    ok, so we have n cosets of Z(G). each one of these corresponds to an inner automorphism of G, that is, we have n distinct possible ways (at most) to conjugate any element of G.
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    Super Member Bernhard's Avatar
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    Re: Centre of a Group and Conjugacy Classes

    I have just read about the Class Equation this morning.

    I am assuming that you are indicating that the proof relies on the Class Equation (thanks for the guidance!)

    I will re-read Dummit and Foote on the Class Equation and apply to the caser when Z(G) has index n

    Peter
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    Super Member Bernhard's Avatar
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    Re: Centre of a Group and Conjugacy Classes

    Thanks

    I think I can see how conjugation is (or leads to) an automorphism of a group - but having difficulty seeing how a coset of Z(G) corresponds to an inner automorphism

    Must go an read next section of Dummit and Foote - ie section 4.4 Automorphisms

    By the way, are you aware of a way to prove this without recourse to an argument re automorphisms?

    Peter
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    Re: Centre of a Group and Conjugacy Classes

    ok, let's just write Z, for Z(G), just for notational purposes.

    suppose that for all g in G, xgx^-1 = ygy^-1. then y^-1xg = gy^-1x, that is, y^-1x is in Z, so Zx = Zy.

    on the other hand, if Zy = Zx, then y = zx for some z in Z, so ygy^-1 = (zx)g(zx)^-1 = z(xgx^-1)z^-1.

    but z commutes with all of G, so it commutes with xgx^-1, so ygy^-1 = z(xgx^-1)z^-1 = (xgx^-1)zz^-1 = xgx^-1.

    that is, all elements of Zx give rise to the same conjugate of g.

    since [G:Z] = n, we can have at most n conjugates of g, one for each coset Zx (it might be that we have considerably fewer,

    as there is nothing to stop ygy^-1 equalling xgx^-1 even when Zx is not Zy).
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    Super Member Bernhard's Avatar
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    Re: Centre of a Group and Conjugacy Classes

    Thanks for that help

    Now working through this.

    Peter
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    Re: Centre of a Group and Conjugacy Classes

    you'll understand Deveno's argument better if you use maps: let a \in G and suppose that A is the conjugacy class of a. let Z be the center of G. define the map \phi : G/Z \longrightarrow A by \phi(gZ})=gag^{-1}. this map is well-defined because if g_1Z=g_2Z, then g_1^{-1}g_2 \in Z and so g_1^{-1}g_2a=ag_1^{-1}g_2, which gives you g_1ag_1^{-1}=g_2ag_2^{-1}. now, obviously \phi is onto and thus n=|G/Z| \geq |A|.
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