Center of a group with finite index

[mahler1]

Homework Statement

Let ##G## be a group such that its center ##Z(G)## has finite index. Prove that every conjugacy class has finite elements.

Homework Equations

The Attempt at a Solution

I know that ##[G:Z(G)]<\infty##. If I consider the action on ##G## on itself by conjugation, each conjugacy class is identified with an orbit, and for each orbit ##\mathcal0_x \cong G/C_G(x)##, where ##C_G(x)## is the stabilizer of ##x## by the action, in this particular case, the centralizer of ##x##. I got stuck here, I know that ##Z(G) \leq C_G(x)## for all ##x \in G##, I don't know how to deduce from here that ##[G:C_G(x)]<\infty##, I would appreciate some help.

 

[pasmith]

Hint: Show that if [itex]g_1[/itex] and [itex]g_2[/itex] are in the same coset of [itex]Z(G)[/itex] then for all [itex]h \in G[/itex] we have [itex]g_1hg_1^{-1}= g_2hg_2^{-1}[/itex].

(Edit: the converse also holds, but I don't think you will require that.)

Given any [itex]h \in G[/itex] you can then show that [itex]\phi_h : G/Z(G) \to [h] : gZ(G) \mapsto ghg^{-1}[/itex] is a surjection, where [itex][h][/itex] is the conjugacy class of [itex]h[/itex]. (The first hint allows us to define such a function.)