Let $G$ be a group and $N \trianglelefteq G$ a normal subgroup. Consider the lower central series

$$G = \gamma_0 G \trianglerighteq \gamma_1 G \trianglerighteq ... \trianglerighteq \gamma_n G= { 1 }$$

terminating in the trivial subgroup after finitely many steps, where $\gamma_{i+1}G := [\gamma_iG, G]$.

How to see that following equation holds:

$$\gamma_i (G/N) = (\gamma_i G)N / N$$

Using induction on $i$ I get only $$\gamma_{i+1} (G/N) = [\gamma_i (G/N), G/N] = [(\gamma_i G)N / N, G/N] \subset [\gamma_i (G), G]N/N = (\gamma_{i+1} G)N / N$$

but how to get an inclusion in other direction?

  • $\begingroup$ You should change the $\zeta$ into $\gamma$, otherwise you are mixing the standard notation for upper and lower central series. $\endgroup$ – Nicky Hekster Feb 26 '18 at 9:47
  • 1
    $\begingroup$ The inclusion in the middle looks like an equality to me. A commutator $[a,b]N$ in $[\gamma_i(G),G)]N/N$ and be written as $[aN,bN]$. $\endgroup$ – Derek Holt Feb 26 '18 at 10:30

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