# The Schur Theorem

The Schur Theorem: if $\left\vert G:Z\left( G\right) \right\vert$ is finite, then $G^{\prime }$ is finite.

My question is: if $1\not=N\trianglelefteq G$ such that $\left\vert G:NZ\left( G\right) \right\vert$ is finite, is there some information on $G^{\prime }$ or

some finiteness conditions involving $G^{\prime }?$

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Well any such condition would have to include $N$ somewhere. It would follow from the quoted result of Schur that $G^{\prime}/(G^{\prime} \cap N)$ is finite. On the other hand, taking $N = G$ tells you nothing really. – Geoff Robinson Aug 15 '12 at 22:17

If $N=G$ this does not give you any information to draw a conclusion on finiteness. You can approach this from another angle.
Theorem If $|G: \zeta_i(G)|$ < $\infty$ then $|G^{i}|$ < $\infty$. Here $\zeta_i(G)$ is the i-th term of the upper central series of $G$ (by convention $Z(G) = \zeta_1(G)$) and $G^{i}$ is the i-th term of the derived series of $G$.
Proof: induction on $i$, let me give you the first step: $G/\zeta_2(G) \cong (G/Z(G))/Z(G/Z(G))$ is finite, implies $(G/Z(G))’$ is finite. But $(G/Z(G))’ \cong G’/(G’ \cap Z(G))$. And of course $G’ \cap Z(G) \subseteq Z(G’)$. We conclude that $|G’/Z(G’)|$ < $\infty$. Again Schur’s Theorem gives you that $G''$ is a finite group.$\Box$
I just looked it up and remembered it when writing down the above Theorem. Reinhold Baer proved something stronger in 1945, namely that if $|G/\zeta_i(G)| < \infty$, then $|\gamma_{i+1}(G)|<\infty$, that is the $(i+1)$-th term of the lower central series of $G$. Note that $G^i \subseteq \gamma_{i+1}(G)$. So the result of Baer implies the Theorem! Conversely, Philip Hall proved that if $|\gamma_i(G)| < \infty$, then $|G/\zeta_{2i}(G)| < \infty$! – Nicky Hekster Aug 16 '12 at 21:38