# A finitely generated group with a finite derived group.

This is a problem from Theory and Problems of Group Theory by B.Baumslag and B.Chandler, McGraw-Hill, 1968, belonging to the Schaum's Outline Series. It is problem 7.57, page 244: $$G$$ is a finitely generated group every element of which has only a finite number of conjugates. Prove that $$G'$$, the derived group, is finite. (Hint: $$\cap C(g_i)=Z(G)$$ where the intersection is taken from $$i=1$$ to $$i=n$$ if $$g_1, ..., g_n$$ are the generators of $$G$$.)

If I could show that $$Z(G)$$ has finite index then by theorem 7.8 in the book $$G'$$ is finite. If $$C(g_i)$$ is of finite index then the intersection, $$Z(G)$$ if of finite index too.
Also if $$G'$$ is finitely generated and every element of $$G'$$ is of finite order then $$G'$$ is finite. Now $$C(g)= {x \in G: xgx^{-1}=g}$$ and for $$g$$ fixed there is a finite number of $$xgx^{-1}$$. This is all I can see. How can I use the hint?

Consider the action of $$G$$ on itself by conjugation. By the orbit-stabilizer theorem, the index of the centralizer (of each generator), $$C(g_i)$$, is the number of conjugates (of that generator). Hence each $$C(g_i)$$ has finite index.
I don't know what I should do: click 'Answer your question' or 'Add a comment'. Anyways thanks a lot. I proved $$C(g)$$ has finite index in two ways. First using the action of $$G$$ on $$G$$ by conjugation and then directly proving that $$f: x^{-1}gx --> xC(g)$$ for fixed $$g$$ is a bijection.