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I am wondering whether there is an easy example of a finite group $G$ with a Sylow $p$-subgroup $P$ and a subgroup $Q\leq P$ such that the normalizer $N_P(Q)$ of $Q$ in $P$ is NOT a Sylow $p$-subgroup of $N_G(Q)$, the normalizer of $Q$ in $G$.

Obviously when trying to find an example, you should look at non-commutative groups. Hence semi-direct products look like a good place to start. So I tried a couple of permutation groups and dihedral groups but failed to find an example. If I remember correctly I found out that such a group should have order at least 24.

To me this problem is not so important, but I would like to see an example. (Which according to some text I came across should be obvious).

Thanks in advance!

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$G=S_4$, $P = \langle (1,2), (1,3)(2,4) \rangle$, $Q = \langle (1,3)(2,4) \rangle$ (order $2$), $N_G(Q) = \langle (2,4), (1,2)(3,4) \rangle$ (order $8$), $N_P(Q) = \langle (1,2)(3,4),(1,3)(2,4) \rangle$ (order $4$).

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  • $\begingroup$ Thank you, I'm surprised I didn't find that myself, I did look at $S_4$. Is there some criterion determining whether this property holds or not? Intuitively this should tell you something about the non-commutativity of the group at hand. I haven't given it much thought though, I'm not really at home in group theory. $\endgroup$ – Mathematician 42 Feb 9 '16 at 10:38
  • $\begingroup$ I don't know of any criterion. generally, as the structure of the group and its Sylow subgroups gets more complicated, the condition becomes less and less likely to hold. $\endgroup$ – Derek Holt Feb 9 '16 at 10:41

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