TITLE: Sylow Subgroups
QUESTION [20 upvotes]: I had been looking lately at Sylow subgroups of some specific groups and it got me to wondering about why Sylow subgroups exist.  I'm very familiar with the proof of the theorems (something that everyone learns at the beginning of their abstract algebra course) -- incidentally my favorite proof is the one by Wielandt -- but the statement of the three Sylow theorems still seems somewhat miraculous.  What got Sylow to imagine that they were true (especially the first -- the existence of a sylow subgroup)?  Even the simpler case of Cauchy's theorem about the existence of an element of order $p$ in a finite subgroup whose order is a multiple of $p$ although easy to prove (with the right trick) also seems a bit amazing.   I believe that sometimes the hardest part of a proving a theorem is believing that it might be true.  So what can buttress the belief for the existence of Sylow subgroups?

REPLY [3 votes]: An extension of the Vipul's ideas can be found in the article (couldn't find a link to the pdf with google)
Subgroup complexes by Peter Webb,
pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987).
But as Mariano already commented, the analogy to the maximal unipotent subgroups of the general linear group was probably not Sylow's motivation. As commented before, he was maybe looking for maximal $p$-subgroups (i.e., maximal with respect to be a $p$-subgroup).
This is also the leitmotif of my favorite proof of the Sylow theorems given by Michael Aschbacher in his book Finite Group Theory. It is based on Cauchy’s theorem (best proved using J. H. McKay’s trick to let $Z_p$ act on the set of all $(x_1, \dots, x_p) \in G^p$ whose product is $1$ by rotating the entries) and goes essentially like this:
The group $G$ acts on the set $\mathrm{Syl}_p(G)$ of its maximal $p$-subgroups by conjugation. Let $\Omega$ be a (nontrivial) orbit with $S\in\Omega$. If $P$ is a fixed point of the action restricted to $S$ then $S$ normalizes $P$ and $PS=SP$ is a $p$-group. Hence $P=S$ by maximality of both $P$ and $S$, and $S$ has a unique fixed point. As $S$ is a $p$-group, all its orbits have order $1$ or a multiple of $p$, in particular $|\mathrm{Syl}_p(G)| = 1 \bmod p$. All orbits of $G$ are disjoint unions of orbits of $S$ proving $\Omega = 1 \bmod p$ and $\Omega' = 0 \bmod p$ for all other orbits $\Omega'$ of $G$. This implies that $\Omega = \mathrm{Syl}_p(G)$, as $\Omega$ was an arbitrary nontrivial orbit of $G$, showing that the action of $G$ is transitive. The stabilizer of $S$ in $G$ is its normalizer $N_G(S)$, and as the action is transitive $|G:N_G(S)| = |\mathrm{Syl}_p(G)| = 1 \bmod p$. It remains to show that $p$ does not divide $|N_G(S):S|=|N_G(S)/S|$. Otherwise, by Cauchy’s theorem there exists a nontrivial $p$-subgroup of $N_G(S)/S$ whose preimage under the projection $N_G(S) \to N_G(S)/S$ is $p$-subgroup properly containing $S$ contradicting the maximality of $S$.