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Let $G$ be a a finite group and $S$ a subnormal subgroup of $G$. The lenght of a fastest chain of subgroups $(U_i)_{1\le i\le n}$ such that $U_1=S$, $U_i$ normal in $U_{i+1}$ and $U_n=G$ is called the defect of subnormality of $S$. Examples in $S_4$ Show that the series of repeated normalizers can be stable and is not reaching the whole group for a subnormal subgroup. But for nilpotent groups this series reaches $G$. My question: Is the length of this series exactly the defect of subnormality for nilpotent groups?

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  • $\begingroup$ This is not true. I have received a counterexample from Mr Stonehewer. $\endgroup$ Apr 16, 2019 at 19:00

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Let H be a cyclic group of order 2 and let K be the dihedral group of order 8 with y a non-central element of order 2. Then let G be the wreath product of H by K with base group B. Then $J = \langle H,Hy\rangle$ is normal in B which is normal in G. But the normalizer of J in G is $B\cdot \langle y \rangle$, which is not normal in G.

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