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 3subgroup and K a Sylow 5subgroup. 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.
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 3subgroup 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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