Let $p$ be a prime factor of the order of a finite group $G$. If $H$ is a normal subgroup $G$ whose index is not a multiple of $p$, show that $H$ must contain every Sylow $p$-subgroup of $G$.
2 Answers
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Hints:
- Show that $H$ must contain a Sylow $p$-subgroup $P$ of $G$.
- Show that $H$ contains all the subgroups of $G$ that are conjugates of $P$.
answered May 12, 2013 at 9:25
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$\begingroup$ Thanks but this is exactly what I want to follow. but I don't know how to go through both of these $\endgroup$– rubinCommented May 12, 2013 at 9:31
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1$\begingroup$ For 1, consider a Sylow $p$-subgroup of $H$. $\endgroup$ Commented May 12, 2013 at 9:56
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$\begingroup$ @rubin: The idea underlying step 1 is that in this case a Sylow $p$-subgroup of $H$ is also a Sylow $p$-subgroup of $G$. Can you see why? $\endgroup$ Commented May 13, 2013 at 3:55
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Hint: If $N$ is normal in $G$, then $N$ contains every element of $G$ that has order coprime to $[G:N]$.
