I'm trying to work though the details of the Sylow theorems. I noticed that the approach taken in Pinter is ordered very different than it is in Aluffi/Hungerford.
This is a lemma I had found myself stuck on. I wanted to run it by people here to make sure I wasn't making any mistakes.
Lemma: Let $G$ be a finite group and $S$ be a maximal $p$-subgroup in $G$. Then $[G : N(S)] \equiv 1 \pmod{p}$.
Proof: Let $S$ act on the set $X$ of conjugates of $S$ in $G$ by conjugation.
Since $S$ is a $p$-group, we have previously established that any action on a set $X$ will give us $|X| \equiv |Z| \pmod{p}$, where $Z$ is the set of fixedpoints. Since $|X| = [G : N(S)]$, it suffices to show that there is exactly one fixedpoint.
Suppose $g \in G$ and $g^{-1}Sg$ is a fixedpoint of this action. Then we have $s^{-1}g^{-1}Sgs = g^{-1}Sg$, which gives us $gs^{-1}g^{-1}Sgsg^{-1} = S$. This puts $gsg^{-1}$ in $N(S)$.
We now appeal to another previously proven lemma that $S$ contains all elements of $p$-power order in $N(S)$. Evidently $gsg^{-1}$ has $p$-power order, since $s$ does. So $gsg^{-1}$ is shown to be in $S$.
This holds for all elements $s \in S$, so we get that $gSg^{-1} \subseteq S$, and since $G$ is finite, we get equality: $gSg^{-1} = S$. Finally, moving the $g$'s to the other side, we get $S = g^{-1}Sg$, and we conclude that the only fixedpoint was $S$ itself.
Does that sound right? I am a little worried, since it does not rely on the first or second Sylow theorems, but yet this seems to be essentially the third Sylow theorem.
The idea for this approach is the one taken by Pinter. But his language is very simple (he makes no use of group actions and relies heavily on "coset-based reasoning") and it's hard to untangle.