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Thread: p-sylow subgroup

  1. #1
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    p-sylow subgroup

    Let $\displaystyle o(G)=pq$, where $\displaystyle p>q$ are primes.
    Since $\displaystyle p$ divides $\displaystyle o(G)$, $\displaystyle G$ contains an element $\displaystyle b$ of order p.
    Let $\displaystyle S=<b>$.
    Since $\displaystyle S$ is a $\displaystyle p$-sylow subgroup of $\displaystyle G$, the number of conjugates is $\displaystyle 1+kp$ for some $\displaystyle k \geqslant 0$.
    But $\displaystyle 1+kp=[G:N(S)]$ which divides $\displaystyle o(G)=pq$.
    Since $\displaystyle (1+kp,p)=1$, then $\displaystyle 1+kp$ divides $\displaystyle q$.
    Since $\displaystyle q<p$, then $\displaystyle k=0 $ and $\displaystyle S$ is normal in $\displaystyle G$.

    I don't understand why S is normal in G. If it is because [G:N(S)]=1, why?
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  2. #2
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    Sylow's theorem explains that. You know that p-sylow groups are all conjugate, so if there is only one p-sylow S in G, then it is normal.
    The reason is that for all x in G, xSx^(-1) is a p-sylow too (his order is p), so it is S.
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