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Here a question I thought of, but can't find an answer to

Find two distinct Sylow $p$-subgroups (of a given $p$) $H_1$ and $H_2$ of $G$ such that $N_G(H_1) = N_G(H_2)$.

I don't know if it's actually possible, so I should qualify the question with

If no such pair exists, show why.

Well, the easiest case would be if $H_1$ and $H_2$ were normal, however this would imply that $n_p = 1$. Hence, we'd only have one Sylow $p$-subgroup.

My intuition says that such a pair does exist, however. Of course, it's merely intuition...

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    $\begingroup$ This is not possible. The normalizer of a Sylow $p$- subgroup has a unique Sylow $p$-subgroup. Well, because the one it has is normal, that is $H_1\unlhd N_G(H_1)$ :-) $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 20:06
  • $\begingroup$ A bit more seriously. This argument is often seen when proving that $N_G(N_G(P))=N_G(P)$ for all Sylow subgroups $P$ of $G$. $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 20:07
  • $\begingroup$ @JyrkiLahtonen Please move your comment to an answer so that this question does not show as unanswered. Thanks. $\endgroup$ – Stephen Meskin Dec 11 '17 at 20:54
  • $\begingroup$ @StephenMeskin James already gave that as an answer (obviously he arrived at it independently from me), but then deleted his answer. I could use my powervote undelete, but I'm a bit hesitant to do that against his wish. $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 21:28
  • $\begingroup$ @JyrkiLahtonen If you won't post the answer, I will as a community wiki. $\endgroup$ – Stephen Meskin Dec 11 '17 at 21:34
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According to a comment by @JyrkiLahtonen and others:
"This is not possible. The normalizer of a Sylow $p$- subgroup has a unique Sylow $p$-subgroup. Well, because the one it has is normal, that is $H_1⊴N_G(H_1)$."

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