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The question is:

There's a group G, with order pm, where p is a prime number and mcd(p,m) = 1. We suppose that G has an unique p-Sylow subgroup P. Proof that P is a normal subgroup of G.


How I understand Sylow's First Theorem, this is obvious from the definition? I've always struggled with giving formal proves, maybe someone can help me in this case, thanks!

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  • $\begingroup$ Obvious from what definition? $\endgroup$ – whacka Apr 9 '15 at 21:39
  • $\begingroup$ You need to check $pPq^{-1}= P$, while $qPq^{-1}$ is a Sylow p group. $\endgroup$ – user99914 Apr 9 '15 at 21:39
  • $\begingroup$ Sorry, I meant Theorem, Sylow's first one! $\endgroup$ – rambrero Apr 9 '15 at 21:40
  • $\begingroup$ Sylow I says there exists a subgroup of order $p$. In general the subgroup guaranteed by Sylow I is not normal. So then, how do you conclude it is normal? $\endgroup$ – whacka Apr 9 '15 at 21:40
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    $\begingroup$ Got it, thanks! $\endgroup$ – rambrero Apr 9 '15 at 21:50
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Sylow's first theorem says that any $p$-group is contained in a $p$-Sylow subgroup. If $P$ is a unique Sylow $p$-subgroup, and if $(p,m)=1$, with $|G| = pm$, then all conjugates of $P$(in fact all elements of order $p$) is contained within $P$, therefore $P$ is normal.

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    $\begingroup$ Thanks a lot, wish I would struggle less with these proofs... $\endgroup$ – rambrero Apr 9 '15 at 21:51
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In fact, you do not need to suppose that $P$ is a $p$-Sylow. If $G$ has a unique subgroup $H$ of a given order, then for any $g \in G$, $gHg^{-1}$ is a subgroup of the same order. Therefore, by uniqueness of $H$, $gHg^{-1} = H$ for all $g \in G$ and $H$ is normal.

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  • $\begingroup$ Thank you, I should have come up with this on my own! $\endgroup$ – rambrero Apr 9 '15 at 21:52
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Hint: For any $q \in G$, the set $qPq^{-1}$ is a subgroup of $G$.

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