Intuition behind picking group actions and Sylow A common strategy in group theory for proving results/solving problems is to find a clever group action. You take the group you are interested in (or perhaps a subgroup), and find some special set that your group can act on, usually by left multiplication or conjugation. Somehow studying this action makes the result you want clearer to see.
The most obvious example of this is with proving the Sylow theorems.

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*To prove the First Sylow theorem (that a $p$-subgroup exists) you can let $G$ act by left multiplication on all subsets of $G$ that have size $p^n$.

*To prove the Second Sylow theorem (that Sylow $p$-subgroups are conjugate) you let $Q$ be any $p$-subgroup, $P$ a Sylow $p$-subgroup, and let $Q$ act on $G/P$ by left multiplication.

*To prove the Third Sylow theorem (that the number of Sylow $p$-subgroups is $1\pmod p$ and divides $|G|$) you can consider both $G$ and some Sylow $p$-subgroup $P$ acting on $\operatorname{Syl}_p(G)$ by conjugation.

My question: is there any intuitive or natural way to come up with these group actions? To me, all three proofs seem magical -- if someone had told me which group action to consider, I could probably have completed the proofs myself, but I would never have come up with the appropriate action myself.
More generally, are there any ways to "see" which group action might help for solving a specific problem?
 A: 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.

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*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 $P \in \text{Syl}_{p}(G)$, $\text{Stab}(P) = N_{G}(P)$.
A: 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.
