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March 19th, 2014, 11:56 AM   #1
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Sylow Theorem

Let G be a group, H a Sylow 3-subgroup and K a Sylow 5-subgroup. Suppose o(H) = 3, o(K) = 5. (The orders of H and K - that is my preferred notation.)

Suppose 3 divides o(N(K)). Then prove that 5 divides o(N(H)).

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April 1st, 2014, 08:28 AM   #2
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Let $\displaystyle k\in K$, $\displaystyle k\ne e$. Then for all $\displaystyle h\in H$, $\displaystyle khk^{-1}$ has the same order as $\displaystyle h$, namely 3. Thus $\displaystyle khk^{-1}\in H$ as $\displaystyle H$ contains all elements of order 3. Therefore $\displaystyle kHk^{-1}=H$ $\displaystyle \implies$ $\displaystyle k\in N(H)$. Hence the normalizer of $\displaystyle H$ contains an element of order 5, namely $\displaystyle k$; the result follows by Lagrange.
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April 8th, 2014, 03:51 AM   #3
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Apologies – the proof I gave above is all wrong! I must do it all over (and it turns out to be a much tougher nut to crack than I thought ).

Now $N(K)$ has an element of order 3 (Cauchy’s theorem) and hence a cyclic subgroup of order 3, $S$. There is a theorem that the centralizer $C(K)$ is a normal subgroup of the normalizer $N(K)$ and $C(K)/N(K)$ is isomorphic to the automorphism group $\mathrm{Aut}(K)$. Since $K$ is cyclic of order 5, we have $|\mathrm{Aut}(K)|=4$, i.e. $|N(K):C(K)|$ is even. $\therefore$ $3\mid |C(K)|$ $\implies$ $S\subseteq C(K)$. But for all $s\in S$, $k\in K$, $sks^{-1}=k$ ($s$ centralizes $k$) $\iff$ $s=ksk^{-1}$ ($k$ centralizes $s$) $\iff$ $K\subseteq C(S)$.

Now $S$ being a Sylow 3-subgroup is conjugate to $H$: there exists $g\in G$ with $S=gHg^{-1}$. Thus $K\subseteq C(gHg^{-1})=g^{-1}C(H)g$ $\implies$ $gKg^{-1}\subseteq C(H)\subseteq N(H)$. Therefore $N(H)$ contains a subgroup of order $|gKg^{-1}|=|K|=5$. The result follows by Lagrange.
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