(Schur) Suppose Z(G) is of finite index in G, then the derived subgroup of G is finite.
We know Schur's lemma that says:
Let |G:Z(G)|=m. Then the map g to g^m is a homomorphism from G into Z(G).
We can establish there are only finitely many commutators in G, and so the derived subgroup G' is finitely generated. Also, since Z(G) is abelian, we see that G' lies in the kernel of the map mentioned above, and hence the m'th power of every element of G' is the identity.
And we know from Schreier's lemma that, if H is a subgroup of G of finite index, where G is finitely generated, then H is finitely generated.
In M.Isaac's algebra book this hint is given. Use Schur's lemma and the following discussion above, and apply Schreier's lemma to G' to show that G'⋂Z(G) is finite.
My question is how to say this and after proving that G'⋂Z(G) is finite, why should G' be finite ?
$$G'/(G'\cap Z(G))\cong G'Z(G)/Z(G)\le G/Z(G)$$
As the last group is finite by hypothesis, the first one above is finite as well, and from here that $G'\cap Z(G)\le Z(G)$ is a finitely generated abelian group and of finite exponent and, thus, finite itself.
Finally, $\;G'\;$ is a finite extension by a finite group and thus finite itself.