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If $G$ is a finite $p$-group and $H$ is a proper subgroup of $G$, then, it is well known that the union of conjugates of $H$ in $G$ is proper subset of $G$. The problem I considering is the following:

Question Let $G$ be a finite $p$-group, $H$ be a non-normal subgroup, such that its normalizer $N_G(H)$ is normal in $G$ (hence, all conjugates of $H$ are contained in $N_G(H)$). Is the union of all conjugates of $H$, a proper subset of $N_G(H)$?

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Just to emphasize this is a property of $p$-groups: If $G$ is not assumed nilpotent, then $G=A_4$ is a counterexample, with $H$ of order 2. Then $N_G(H)=K_4$ is the union of the three $G$-conjugates of $H$. –  Jack Schmidt Jul 23 '13 at 15:23

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I believe the answer is yes, the union of the conjugates of $H$ is a proper subset of $N_G(H)$.

Let $H < G$ be a counterexample with $|G|$ minimal. Since $N_G(H) \lhd G$, the intersection $Z := N_G(H) \cap Z(G)$ of $N_G(H)$ with the centre of $G$ is nontrivial.

If $Z < H$, then $H/Z < G/Z$ is a smaller counterexample, whereas if $Z \not< H$ then there exists $g \in Z \setminus H$ and then $g$ is not contained in any conjugate of $H$.

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This question considers the case of $H$ being 2 steps subnormal. Since a subgroup of a $p$-group must be subnormal, it is natural to ask: Is there some generalization of this to $k$ steps subnormal $H$? For example, what if $N_G(N_G(H)) \lhd G$? –  Pablo Jul 8 at 8:17

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