# Thread: Centre of a Group and Conjugacy Classes

1. ## 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

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

Peter

2. ## 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.

3. ## Re: Centre of a Group and Conjugacy Classes

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

4. ## 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

5. ## 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).

6. ## Re: Centre of a Group and Conjugacy Classes

Thanks for that help

Now working through this.

Peter

7. ## 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 $n=|G/Z| \geq |A|$.