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:

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

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

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.

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

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

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).

Thanks for that help

Now working through this.

Peter

you'll understand Deveno's argument better if you use maps: let and suppose that is the conjugacy class of . let be the center of . define the map by this map is well-defined because if , then and so which gives you now, obviously is onto and thus .