Intuition behind picking group actions and Sylow

Question

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.

1. 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 $p^n$.
2. 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.
3. 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?

Answer 1

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

Answer 2

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.

Answer 3

I wrote a blog post Meditation on the Sylow theorems which among other things is an attempt to understand this question. I also saw the proof of Sylow I via the action on subsets of size $p^n$ as an undergraduate and found it too magical and mysterious; I still find it too magical and mysterious, so the linked post presents several other proofs of Sylow I which avoid it.

Here are sketches of the proofs. For proofs 1 and 2, an argument due to Frobenius shows that if a group $H$ has a Sylow $p$-subgroup $P$ (by which I mean a $p$-subgroup of the maximal order allowed by the order of $H$), then so does every subgroup $G$ of it. The proof involves considering the action of $G$ on $H/P$ by left multiplication. I think this is pretty intuitive: the point is that by hypothesis $|H/P|$ is not divisible by $p$, so there must be some orbit of the action of $G$ whose size is also not divisible by $p$, and this turns out to imply the result. This is a pretty intuitive way to use the hypothesis that $P$ is Sylow, I think.

This argument reduces Sylow I to finding, for every group $G$, a way to embed it into a larger group $H$ for which you can explicitly write down a Sylow $p$-subgroup.

1. Proof 1 embeds $G$ into a symmetric group $S_n$, then explicitly constructs the Sylow $p$-subgroups of $S_n$; as far as I know this was first done by Cauchy (to prove Cauchy's theorem!).

2. Proof 2 embeds $G$ into a general linear group $GL_n(\mathbb{F}_p)$, whose Sylow $p$-subgroup is the group of unitriangular matrices. Serre presents this proof in a textbook; I don't know who it's due to.

3. Proof 3 is the subsets of size $p^n$ proof, which Serre attributes to Miller-Wielandt; I try to motivate it a bit in the linked post but overall I think it is both the hardest argument to motivate and the one you learn the least from.

4. Proof 4 is an argument by induction on $|G|$ using the abelian case and the class equation; it is apparently due to Frobenius. I like this argument because it's more explicit than the others and in particular gives an algorithm for constructing Sylow subgroups.

5. Proof 5 is Sylow's proof! It actually proves that for every prime power $p^k \mid |G|$, $G$ has a subgroup $P$ of size $p^k$. The argument is by induction on $k$, so the base case $k = 1$ is Cauchy's theorem, and the inductive step uses normalizers and the action of $P$ on $G/P$. I also like this argument for being explicit, and it gives a (different) algorithm for constructing Sylow subgroups.

As for the question of how to come up with group actions, once you eliminate the subsets-of-size-$p^n$ construction which I think is harder to motivate than the others, the remaining ideas are not so bad. The Sylow theorems are true for arbitrary finite groups, so the proofs must involve constructions which one can perform on an arbitrary finite group $G$. There are not so many of these: $G$ acts on itself by left multiplication, it acts on itself by conjugation, it acts on its subgroups by conjugation, it acts on $G/H$ for any subgroup $H$, and all of these actions can be restricted to subgroups. That's already enough.

The motivation for wanting to specifically act with a $p$-subgroup $P$ on things is the preposterously powerful $p$-group fixed point theorem $|X^P| \equiv |X| \bmod p$, which I also wrote about here. The Sylow proofs frequently involve applying this result; I'm amazed it doesn't appear to have its own standard name so I want to popularize this one, which seems like the most intuitive option.

Answer 4

Synthesizing some of the other answers, I think many of these "magical" group actions come from just trying to rewrite common group theoretic concepts in the orbit-stabilizer language. (The fact that one can do so, is in my opinion an unexpected small miracle.)

For instance (as I learned from this MSE answer Orbit stabiliser theorem as an analogue to first isomorphism theorem), the kernel and image of the group homomorphism $\phi: G \to H$ can be thought of (resp.) as the stabilizer and orbit of $e_H$ under the group action $G \curvearrowright H$ defined by $g \cdot h := \phi(g) h$. Then the orbit-stabilizer theorem gives us precisely the 1st isomorphism theorem.

The famous $p$-group fixed point theorem that Qiaochu enthuses over essentially promises us non-trivial fixed points (in fact a decent number of them; in some applications it's furthermore important to know that there's a multiple-of-$p$ many total fixed points). So, in fact we can take advantage of this powerful theorem, if only we could manhandle important concepts into being related to fixed points of some action.

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Example: the normalizer of a subgroup $H$ in a group $G$. Obviously, $$g\in \text{N}(H) \iff g^{-1}Hg=H$$ it is instantaneous to see the normalizer as a stabilizer: $g\in \text{N}(H) \iff g\in \text{stab}_G(H)$ w.r.t the action $G \curvearrowright 2^G$ (all subsets of $G$) given by $g \mathbin{‣} S := g^{-1} Sg$.

But a slightly further meditation on the definition $$g\in \text{N}(H) \iff g^{-1}hg \in H \text{ for all } h\in H \iff hgH = gH \text{ for all } h\in H $$ reveals the normalizer as the fixed points: $g\in \text{N}(H) \iff gH\in \text{fix}((G/H) \curvearrowleft H)$ w.r.t the action $H \curvearrowright G/H$ (all left cosets of $H$) given by $h \cdot xH := h xH$.

Somehow, the 2nd perspective is "more naturally group theoretic" than the first "more obvious" perspective; the first one has a group action on the power set of $G$ (a set theoretic notion, not a group theoretic notion), but the 2nd one has a group action on left cosets (indeed a group theoretic notion). A little mysterious.

Conclusion: $\text{fix}((G/H) \curvearrowleft H) = \text{N}(H)/H$ (as sets).

If $H$ is a $p$-group, the $p$-group fixed point theorem tells us that $[G:H] \equiv [\text{N}(H):H] \mod p$, which in particular tells us that the quotient group $\text{N}(H)/H$ has order divisible by $p$, as long as $H$ is not a $p$-Sylow subgroup of $G$.

This forms the basis of the induction step (base step is Cauchy's theorem) in this KConrad writeup of Sylow I.

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Above was about the stabilizers of the action $\mathbin{‣} : G \curvearrowright 2^G$. What about orbits? Two subgroups $H',H$ are conjugate iff $H' \in \text{orb}_G(H) \iff \exists g$ s.t. $g^{-1}H'g = H$. But doing same "meditation" as above, $$g^{-1}H'g = H \iff g^{-1}h' g \in H \text{ for all } h'\in H \iff h'g H = gH \text{ for all } h'\in H.$$ Aha! We recognize again that this is a fixed point equation!

We have shown that $H',H$ are conjugate $\iff$ there is a fixed point to the action $H' \curvearrowright G/H$ (all left cosets of $H$) given by $h' \cdot xH := h' xH$.

This is exactly what leads to Sylow II (as discussed in anon's answer, and also KConrad's writeup)

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Something very interesting I noticed: in Sylow I, we use that $0\equiv [G:H] \equiv [\text{N}(H):H] \mod p$ to show that in fact $\text{N}(H) \supsetneq H$ (i.e. strictly larger, i.e. non-trivial elements of $\text{N}(H)$).

But in Sylow II, we use that $0 \not\equiv [G:P] \equiv \text{fix}((G/P) \curvearrowleft P') \mod p$ to show existence of non-trivial fixed points.

So in both cases, we conclude non-trivial fixed points of some kind, but in one case we get the conclusion by being $\equiv 0 \mod p$ and in the other, we get the same conclusion by being $\not\equiv 0 \mod p$!!! That's kind of crazy!

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Sylow II spoon feeds the correct next action: conjugation on $\mathcal {Syl}_p(G)$, the set of $p$-Sylow subgroups. Start with conjugation $G \curvearrowright \mathcal {Syl}_p(G)$ (which already gives us 2 results, one of which KConrad calls Sylow III* --- see KConrad's table). Then as anon's answer says, is not a far walk to the idea of considering conjugation of just $P \curvearrowright \mathcal {Syl}_p(G)$, and then Sylow III follows.

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Summary: the Sylow theorems pop right out, if you try to rephrase normalizers (and conjugation of subgroups) in terms of fixed points (O glorious $p$-group fixed point theorem!!!)