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Let $G=PSL(2,n)$ where $n$ is odd prime power and $n$ is not a prime. Also $|G|=\frac{1}{2}(n-1)n(n+1)$. We know that $G$ has subgroup $N$ that is a frobenius group with kernel $K$ and complement $H$ where $|K|=q$ , $|H|=\frac{1}{2}(n-1)$. So $K$ is a Hall-subgroup of $G$. Suppose $x\in G$ and $o(x)=p$ where $p$ divides $n$. Is it true that there is a conjugat of $K$ namely $K_1$ such that $x\in K_1$? why?

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You have both $q$ and $n$... are they supposed to be the same? Is $q$ a power of $p$ ? –  Ted Nov 21 '12 at 8:07
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Sylow's theorem! $K$ is a Sylow $p$-subgroup of $G$. –  Geoff Robinson Nov 21 '12 at 8:54
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That is true. Since every $p$-sylow subgroup of $K$ where $p|n$ is a $p$-sylow subgroup of $G$ too, let $P_1$ be a $p$-sylow subgroup of $G$ that $x\in P_1$ and let $P$ be a arbitrary $p$-sylow subgroup of $K$. So theree exist $g\in G$ such that $x\in P_1=^g\leq K^g$.

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thanks for your good answer. –  Adeleh Nov 22 '12 at 10:35
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