Intuition behind picking group actions and Sylow

The Sylow Theorems definitely have a lot of non-obvious ideas, and I think expecting someone to come up with these ideas on their own is not a reasonable ask. With that said, here are some pieces of intuition I have.

  • The Sylow Theorems are inherently combinatorial. It is natural to ask for an underlying combinatorial object (or collection of such objects) upon which the group should act. Let $n := p^{\alpha}m$, where $p$ does not divide $m$. The Lucas Congruence tells us that $p$ does not divide $\binom{p^{\alpha}m}{p^{\alpha}}$. The binomial coefficient tells us that it might be worth looking at the $p^{\alpha}$ subsets of the group. In terms of the action, left multiplication and conjugation are the natural things to try. The left action is a bit easier to analyze; so if I hadn't seen the proof before, I'd start with that.

  • When analyzing a group action, the orbits and stabilizers tell us a lot. Once we have found that the stabilizer of the above action is a Sylow $p$-subgroup, it is very natural to ask whether we can get all of the Sylow $p$-subgroups from the stabilizer. This would actually be very nice if we could do so; as otherwise, we have a lot more structural and combinatorial analysis to do, which can be messy. Note that conjugation gives us automorphisms. So the conjugates of the stabilizer are also Sylow $p$-subgroups, but we don't know if we have enumerated all the Sylow $p$-subgroups.

  • To get that the Sylow $p$-subgroups are conjugate, we have a few observations. First, the $p$-group fixed point theorem that if a $p$-group $P$ acts on a finite set $X$, where $p$ does not divide $|X|$, then the action has a fixed point. We use this fact, together with the intuition that every $p$-subgroup of $G$ should be a subgroup of a Sylow $p$-subgroup to get the conjugation action. Precisely, if $P$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $|G/P|$. Let $Q$ be a $p$-subgroup of $G$. We let $Q$ act on $G/P$ by left multiplication. The conjugacy condition comes from analyzing the fixed points. There are technicalities to this in the proof, but these are the main ideas (at least, to me).

  • The condition that $n_{p} \equiv 1 \pmod{p}$ and $n_{p}$ divides $m$ falls out from the Orbit-Stabilizer Theorem. Once we have that $G$ acts transitively on $\text{Syl}_{p}(G)$ by conjugation, we have that for $P \in \text{Syl}_{p}(G)$, $\text{Stab}(P) = N_{G}(P)$.


Well, there's only a few group actions to consider. The ones that worked made it into the proofs. That said, there are a priori reasons to check these actions used in the proofs first.

In investigating these actions, having a fundamental understanding of orbit-stabilizer is critical.

For Sylow I, we don't know from the get-go there is any subgroup of order $p^n$. But of course there are subsets of size $p^n$, and the subgroups are distinguished among these in that they are closed under multiplication. This closure property can be rephrased in terms of stabilization and fixed points. Indeed, if a subset has stabilizer $S$ (under left multiplication), then it is a union of right cosets of $S$, so in particular stabilizers must be $p$-subgroups. By orbit-stabilizer, a subset has full Sylow stabilizer iff its orbit size is not divisible by $p$. Which suggests a possible way to prove Sylows exist by contradiction by considering the sum of orbit sizes mod $p$!

For Sylow II, recall some basic facts about group actions a la (again) orbit-stabilizer. Every $G$-set is a disjoint union of orbits, all orbits are eqiuvalent to $G/H$ for stabilizers $H$. But these coset spaces are only inequivalent up to conjugacy class of subgroup, because $\mathrm{Stab}(g\omega)=g\mathrm{Stab}(\omega)g^{-1}$. This can be rephrased as follows: $H,K\le G$ are conjugate iff $K$ has a fixed point acting on $G/H$ (by left-multiplication).

For Sylow III, after we already know Sylow subgroups are conjugate we automatically know they carry a conjugation action from the group. Orbit-stabilizer gets you $G/N_G(P)$. A trick we used for Sylow I to get a numerical result is to mod by $p$, so we can do that here to investigate the size of this index, which means we want to consider the action of a $p$-group for modding out to yield results. Might as well a Sylow subgroup, and further might as well $P$ so we have a guaranteed fixed point.