Subgroups of nilpotent groups are subnormal

Question

This lemma is taken from Robinson's A Course in the Theory of Groups. Why does $$\zeta_{i+1}G/\zeta_iG=\zeta(G/\zeta_iG)$$ imply that $H\zeta_iG\triangleleft H\zeta_{i+1}G$? I'll post my answer, if I find one.

Answer 1

First of all, given any group $G$ and a subgroup $H$ one has that $$Z(G)\triangleleft C_G(H)\triangleleft N_G(H)$$ So that in fact $H\triangleleft H\ Z(G)$. Now consider the general case of a group $G$ with a normal subgroup $N$ and a subgroup $K$ containing $N$ such that $ K/N=Z(G/N)$, and take any $H\leq G$: then $$HK/N=HN/N\ HK/N= HN/N\ Z(G/N)$$ thus $HN/N\triangleleft HK/N$ and by the correspondence theorem $HN\triangleleft HK$.

Setting $N=\zeta_i(G)$ the claim above holds.

Answer 2

I try to prove that $HZ_i/Z_i\lhd HZ_{i+1}/Z_i$

Let $abZ_i\in HZ_{i+1}/Z_i$ and $hZ_i\in HZ_i/Z_i$ Then $$abZ_ihZ_i(abZ_i)^{-1}=aZ_ihZ_ia^{-1}Z_i$$ Here uses the fact that $bZ_i\in Z_{i+1}/Z_i=Z(G/Z_i)$.
Hence $$abZ_ihZ_i(abZ_i)^{-1}=aha^{-1}Z_i\in HZ_i/Z_i$$

So $HZ_i/Z_i\lhd HZ_{i+1}/Z_i$ and hence $HZ_i\lhd HZ_{i+1}$

Remark: $Z_i$ stands for $\zeta_i G$