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I am trying to construct an example of a group $G$ and a subgroup $H$ where the 1-hypernormalizer and the 2-hypernormalizer of $H$ are distinct.

The normalizer of $J \leq G$ is $N(J)=\{g \in G \mid gJg^{-1}=J\}$. The k-hypernormalizer of $H$ is the k-th iteration of $N(-)$ applied to $H$.

I tried but was not able to find concrete examples in a literature search (one book that comes up in references but that I unfortunately don't have is 'A Course in the Theory of Groups' by Robinson).

I have thought about finding a non-normal subgroup of a group with the normalizer condition: $J \lneq N(J)$ for all subgroups $J \leq G$. If $G$ is finite, then $G$ satisfies the normalizer condition if $G$ is nilpotent (I think also only if). However, I don't have much experience with such groups, so I am looking for some directions to follow that may be useful.

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Your idea is fine. The easiest nilpotent groups are the dihedral 2-groups.

Let $G_n = \langle f, r : f^2 = r^{2^n} = (fr)^2 = 1 \rangle$ be the dihedral group of order $2^{n+1}$, and for $0 \leq k \leq n$ let $N_k = \langle f, r^{2^{n-k}} \rangle$ be a dihedral subgroup of order $2^{k+1}$. If $1 \leq k \leq n$, then the normalizer of $N_{k-1}$ in $G$ is $N_k$, and so the $k$th hyper normalizer of $J=N_0=\langle f\rangle$ is $N_k$, as long as $k \leq n$.

The restricted direct product of the $G_n$ has all hyper-normalizers distinct (starting with the restricted direct product of the $N_0$, and defining $N_k = G_n$ when $k \geq n$).

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Thank you very much! $D_8$ is a perfect example for what I wanted. –  Stacker Mar 16 '12 at 18:44

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