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In relation to an earlier post I made, Prove there is no element of order 6 in a simple group of order 168

The answer provided relies on this statement "Therefore the normalizer of any Sylow 3-subgroup would have to be cyclic of order 6, and an element of order 6 belongs to exactly one such normalizer"

How does one prove that if one normalizer is cyclic, then they all must be cyclic? I accepted it as obvious at first, but then when I tried to prove it I found myself stuck and unable to show that this must be the case.

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    $\begingroup$ Sylow subgroups are conjugate, so their normalizers will also be conjugate; isn't it? $\endgroup$ – Beginner Oct 14 '11 at 6:21
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    $\begingroup$ One meta way to do this is to ask "Did he use anything about which particular Sylow 3-subgroup it was?" If not, then he not only showed a particular normalizer was cyclic, he proved a theorem about normalizes of Sylow 3-subgroups in that group: they are cyclic. Whenever you see a conclusion, it is a good idea to make a quick note of what the hypotheses of that conclusion are. That quick note is a (possibly useless, but possibly awesome) theorem. $\endgroup$ – Jack Schmidt Oct 14 '11 at 14:23
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Lemma. Let $H<G$ and $g\in G$. Then $N_G(H^g) = (N_G(H))^{g}$.

Proof. If $x\in N_G(H^g)$, then for every $h\in H$ we have $x(ghg^{-1})x^{-1}\in H^g$; hence for each $h\in H$ there exists $h'\in H$ such that $xghg^{-1}x^{-1} = gh'g^{-1}$. Thus, $(g^{-1}xg)h(g^{-1}xg) = h'\in H$. That is, $g^{-1}xg\in N_G(H)$, so $x\in gN_G(H)g^{-1}$.

Thus, $N_G(H^g)\subseteq (N_G(H))^g$.

For the converse inclusion, by the same argument we have $$(N_G(H))^g = \Bigl( N_G\bigl( (H^g)^{g^{-1}}\bigr)\Bigr)^g \subseteq \Bigl(\bigl(N_G(H^g)\bigr)^{g^{-1}}\Bigr)^g = N_G(H^g),$$ proving equality. $\Box$

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Two Sylow subgroups are conjugate. An element which conjugates them also conjugates their normalizers.

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