The normalizer of a subgroup is the join of all subgroups in which it is normal

Let G be a group and H \leq G. Prove that if N is a normal subgroup of H then H \leq N_G(N). Deduce that N_G(N) is the largest subgroup of G in which N is normal (i.e. the join of all subgroups H for which N \leq H is normal).


Note that if h \in H, then hNh^{-1} = N since N is normal in H. Hence H \subseteq N_G(N), so that H \leq N_G(N) by a lemma to a previous exercise.

Now if we consider the set \mathcal{A} = \{ H \leq G \ |\ N \vartriangleleft H \}, note that N_G(N) \in \mathcal{A} and if H \in \mathcal{A} then H \leq N_G(N). Hence N_G(N) is \subseteq-greatest among the subgroups of G in which N is normal.

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