Normalizer of subgroup of a finite $p$-group.

Asked Jul 02 '23 at 21:05

Active Jul 02 '23 at 21:42

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I have been struggling with the following exercise for a while now:

Let $p$ be a prime number and $G$ be a finite $p$-group. Show for any real subgroup $H$ of $G$ that $H$ is a real subgroup of the normalizer $N_G(H)$.

In the case $Z(G)\nsubseteq H$ the solution is simple. I have been especially struggling with the case that $Z(G) \subseteq H$.

Anybody have any hints on what theorems to use?

Thanks in advance <3

edited Jul 02 '23 at 21:42

Shaun

47,747

asked Jul 02 '23 at 21:05

martinr

Don't you have $H \subset N_G(H)$? – julio_es_sui_glace Jul 02 '23 at 21:16
Yes, that's trivial. But I need to show that $H$ is a real subset of $N_G(H)$. So I should find one element $g \in G\setminus H$ such that $gHg^{-1}=H$. – martinr Jul 02 '23 at 21:18
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Ok by real you mean strict or proper – julio_es_sui_glace Jul 02 '23 at 21:21
If would make a disjunction $H$ is a $p$-group and $H$ is not and use that all $p$ syllow are conjugated? – julio_es_sui_glace Jul 02 '23 at 21:23
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What is a "real subgroup"? – Shaun Jul 02 '23 at 21:28
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That was asked here before. See here: (though I don't know why in the answer the action was defined on right cosets. IMO it is more natural to act from the left on left cosets) https://math.stackexchange.com/questions/168963/proper-subgroups-of-finite-p-groups-are-properly-contained-in-the-normalizer – Mark Jul 02 '23 at 21:29

Normalizer of subgroup of a finite $p$-group.

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