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This is the excerpt of the proof of Sylow's theorem from Lang's and some moments of the are unclear to me.

1) In order to prove (ii) they take $H$ to be Sylow $p$-subgroup and it is not obvious to me why any two Sylow $p$-subgroups are conjugate?

2) Is $H$ is a Sylow $p$-subgroup then why it has only one fixed point?

I have spent some hours in order to understand these questions but was not able.

Would be very grateful for detailed help!

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    $\begingroup$ It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$? $\endgroup$ – Eric Wofsey Nov 4 '18 at 22:16
  • $\begingroup$ @EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $H\subset Q$ $\Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate? $\endgroup$ – ZFR Nov 4 '18 at 22:22
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The argument shows that for any $p$-Sylow subgroup $H$, there exists $Q\in S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).

Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.

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    $\begingroup$ Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1 $\endgroup$ – ZFR Nov 4 '18 at 22:31

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