Proof of Sylow's second and third theorem from Lang's book

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!  • 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 $H=Q$? – Eric Wofsey Nov 4 '18 at 22:16
• @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? – ZFR Nov 4 '18 at 22:22

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 $$H$$.