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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?

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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.

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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.

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