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 March 19th, 2014, 11:56 AM #1 Newbie   Joined: Mar 2014 Posts: 17 Thanks: 2 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)). Any suggestions? April 1st, 2014, 08:28 AM #2 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 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. Thanks from Deveno Last edited by Olinguito; April 1st, 2014 at 08:37 AM. April 8th, 2014, 03:51 AM #3 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 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. Tags sylow, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ThatPinkSock Abstract Algebra 1 November 11th, 2011 04:25 AM WannaBe Abstract Algebra 0 December 22nd, 2009 10:35 AM WannaBe Abstract Algebra 0 December 14th, 2009 04:03 AM Kinga Abstract Algebra 1 November 13th, 2007 01:32 AM weier Abstract Algebra 1 November 20th, 2006 10:40 AM

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