Proof of Sylow's theorem. I read this proof of Sylow's theorem in Rotman's "An introduction to the Theory of Groups"  and I don't understand what is the argument in the second paragraph (the one in the green box) for. Isn't the first paragraph enough to say that the proof is finished? They took a Sylow $p$-subgroup $Q$ of $G$ and showed that $Q=P_i$ for some $P_i \in X$, for me that is enough to prove that every Sylow $p$-subgroup of $G$ is a conjugate to
 $P$. After that, in the second paragraph, they use a condtradiction argument to prove the same thing they already proved, don't they? Maybe it's something about logic what is making me confused, I don't know. What do you think?

 A: Yes, this argument is sometimes slightly confusing, so one must be careful. In fact, the book has a typo and the argument is rather poorly explained.
We have all our Sylow subgroups, $X=\{P_1,\ldots,P_n\}$. We already know one exists, which is $P=P_1$. Let $G$ act on $X$ by conjugation. Then $P$ is in some $G$-orbit, call it $\mathcal X_1$. Then $P$ acts on this orbit $\mathcal X_1$ by conjugation too.  Since $P$ is a $p$-group, the orbit stabilizer theorem shows the $P$-orbits have size a power of $p$ (they might have size $p^0=1$), that is, we have $\mathcal X_1$ decomposed as $P$-orbits $\mathcal O_1,\ldots,\mathcal O_k$. Now, suppose an orbit is a singleton, $\{P_i\}$. This means that for any $x\in P$, $xP_i x^{-1}=P_i$; that is $P\subseteq N(P_i)$. But you've probably shown that $P$ is the only $p$-Sylow subgroup of $N(P)$, for example, by showing that if $H$ is a $p$-subgroup of $N(P)$ then $H\leqslant P^1$. This means that the only $P$-orbit of the action of $P$ on $\mathcal X_1$ that has size $1$ is the orbit $\{P\}$. So we've proven that $|\mathcal X_1|=1\mod p$. Now we'll prove that $\mathcal X_1=X$, i.e. that every $p$-Sylow is conjugate to $P$ in $G$.
Suppose for the sake of a contradiction that $Q$ is another Sylow subgroup that is not in $\mathcal X_1$. If we make $Q$ act on $\mathcal X_1$, we see there are no $Q$-orbits of size $1$, since this would mean the $Q$-orbit is $\{Q\}$ by the above, but $Q\notin \mathcal X_1$! We conclude that every orbit has size a true power of $p$; and  $|\mathcal X_1|=0\mod p$, which is absurd by the finding above. Hence $Q\in\mathcal X_1$ for every Sylow subgroup, i.e. $X=\mathcal X_1$.

$1.$ The proof of this is quite interesting on its own: suppose $H$ is a $p$-subgroup of $N(P)$. Then $PH$ is a subgroup of $N(P)$ since $P$ is normal in $N(P)$, $P\lhd PH$ and $$\frac{PH}P\simeq\frac{H}{H\cap P}$$ so that $PH$ is a $p$-group (its order is a power of $p$). But $P\leqslant PH$. Since $P$ is maximal, $PH=P$ so $H\leqslant P$, as we wanted.
