Today I Learned

June 23rd

Today I learned (and reviewed) some proofs of the Sylow theorems that I like. These all come from group actions, and many from the $p$-group fixed-point theorem. To not be too bogged down in details (as I usually am), we often won't show that these group actions are actually actions. That kind of rigor is not the point today.

To avoid digressions in what follows, we review the necessary theory at the start, knowingly sacrificing motivation. Here is the orbit-stabilizer theorem.

We send a coset $g\op{Stab}(s)$ to the element $gs$ in the orbit. This mapping is well-defined because $g\op{Stab}(s)=h\op{Stab}(s)$ if and only if $h^{-1}g\in\op{Stab}(s)$ if and only if $\left(h^{-1}g\right)(s)=s$ if and only if $gs=hs.$

To show the bijection, we note the map is surjective because we can hit $gs\in Gs$ with the coset $g\op{Stab}(s).$ And the map is injective because we showed $gs=hs$ if and only if $g\op{Stab}(s)=h\op{Stab}(s)$ above. This finishes the proof. $\blacksquare$

This gives us the $p$-group fixed-point theorem.

This is done by writing down the class equation. Note that orbits partition our set $S$ because being in the same orbit is an equivalence relation. This lets us write\[\#S=\sum_{Gs\text{ orbit}}\#(Gs).\]If the size of the orbit is $1,$ then $Gs=\{s\}$ is a fixed point, so we can remove all these points and write\[\#S=\op{Fix}(S)+\sum_{\substack{Gs\text{ orbit}\\\#(Gs) \gt 1}}\#(Gs).\]It remains to show that the sum on the right-hand side is divisible by $p$; we show that all terms in the sum are divisible by $p.$

Indeed, the size of the orbit $Gs$ is equal to the number of cosets in $G/\op{Stab}(s),$ which is equal to $[G:\op{Stab}(s)] \gt 1.$ This index divides the size of the total group $G,$ so it is a power of $p.$ Powers of $p$ larger than $1$ are divisible by $p,$ so we are done. $\blacksquare$

A few of our group actions will make the normalizer of a subgroup appear, so it is probably safest to define it at the outset so that we aren't surprised when it comes up.

It is not difficult to show that $N(H)$ is in fact a subgroup. Indeed, it is closed under inversion because $gHg^{-1}=H$ impies $H=g^{-1}Hg,$ and it is closed under multiplication because $g_1Hg_1^{-1}=g_2Hg_2^{-1}=H$ implies\[(g_1g_2)H(g_1g_2)^{-1}=g_1g_2Hg_2^{-1}g_1^{-1}=g_1Hg_1^{-1}=H.\]Anyways, what we like about the normalizer is that $H$ is a subgroup of $N(H)$ ($H$ is closed), and in fact a normal subgroup. Indeed, for any $g\in N(H),$ we see $gHg^{-1}=H,$ which is what we need for $H$ to be normal in $N(H).$

Before we continue with the proofs of the Sylow theorems, we codify the following intuition.

We are now ready to talk about the Sylow theorems. We will prove them in sequence; here is the first Sylow theorem, which asserts existence.

It is not easy, but it is possible, to prove this directly using group actions. We will start with a relatively unmotivated approach and then return to prove this in a more motivated way. It might be productive to not read this proof at all.

Let $S$ be the set of all subsets of $G$ of order $p^\nu.$ The idea here is that we would like to use a group action to "detect'' a subgroup out of $S.$ That is not exactly what is about to happen, but it explains why we might consider such a thing. Because this set $S$ is somewhat exotic, we will have $G$ act on it by left multiplication.

Very quickly, this action is well-defined because multiplication by $g\in G$ is a bijection. Namely, we need to know $\#gA=\#A$ for $A\in S$ to have $gA\in S,$ and indeed, the maps $a\mapsto ga$ and $a\mapsto g^{-1}a$ are inverses of each other and therefore each bijections. We will not say more than this.

Our action is not by a $p$-group, so we cannot use the $p$-group fixed-point theorem, but we can say something similar. Note that\[\#S=\binom{p^\nu m}{p^\nu}\equiv\binom m1=m\not\equiv0\pmod p\]by Lucas's theorem on binomial coefficients; see here or even here for a reference, but we won't prove it now because it would take us too far afield. The point of saying is that when we partition $S$ into orbits\[\#S=\sum_{Gs\text{ orbit}}\#(Gs),\]we cannot have all of the orbits have size divisible by $p.$ So in a style akin to the $p$-group fixed-point theorem, we get to pick up a set $A\in S$ whose orbit $G\cdot A$ has size not divisible by $p.$

We said above that we would like to detect a Sylow $p$-subgroup directly from our action, but we cannot quite do that because $A$ might, for example, be the coset of a Sylow $p$-subgroup by merely living in the orbit of a bona fide Sylow $p$-subgroup. There are coarse ways of dealing with this, but to use $p\nmid\#(G\cdot A),$ we note\[\#(G\cdot A)\cdot\#\op{Stab}(A)=\#G=p^\nu m\]by the orbit-stabilizer theorem. Because $\#A$ has lots of $p$s, it is reasonable (a la Lagrange) to expect $\#(G\cdot A)$ to have few $p$s, and indeed it has no $p$s by construction of $A.$ This forces $\op{Stab}(A)$ to have lots of $p$s, in fact all of them, so $p^\nu\mid\#\op{Stab}(A).$ This indicates that we want to show $\op{Stab}(A)$ is Sylow $p$-subgroup!

So far we have used the size of the group $G$ as well as some number theory to show that $\#\op{Stab}(A)$ is divisible by $p^\nu.$ To show that $\#\op{Stab}(A)$ is actually equal to $p^\nu,$ it remains to use the group action in a meaningful way. We only know what $\op{Stab}(A)$ does on $A,$ namely it will take any $a\in A$ somewhere else in $A.$ This tells us\[\op{Stab}(A)\cdot a\subseteq A\]for a given $a\in A.$ We would like to say that this means $\#\op{Stab}(A)\le\#A,$ and indeed, if $g,h\in\op{Stab}(A)$ with $ga=ha,$ then right multiplication by $a^{-1}$ implies $g=h.$ So the multiplication by $\op{Stab}(A)$ sends $a$ to a different element each time, so we get\[\#\op{Stab}(A)=\#\big(\op{Stab}(A)\cdot a\big)\le\#A=p^\nu.\]Because $p^\nu\mid\#\op{Stab}(A)$ already, it follows $\op{Stab}(A)$ has size $p^\nu$ and is our Sylow $p$-subgroup. $\blacksquare$

Above we said around the middle that a barrier to showing $A$ is the desired Sylow $p$-subgroup is that $A$ might be a coset of a Sylow $p$-subgroup. We say now that the end of our proof shows that this is exactly the barrier: we showed\[\op{Stab}(A)\cdot a\subseteq A,\]and because we established both sides have size $p^\nu,$ it follows $\op{Stab}(A)\cdot a=A.$ Because $\op{Stab}(A)$ was our Sylow $p$-subgroup, we see that, indeed, $A$ was a (right) coset of a Sylow $p$-subgroup all along.

The above proof was overly clever, and though I like it, it is not using any of the more standard group actions I am trying to get used to. So we present a second proof, which proves the result through the following lemma.

This time we want to use the $p$-group fixed-point theorem, and the only $p$-group immediately at our disposal is $H.$ It needs to act on something, and for this proof, it will act by multiplication on its cosets $G/H.$ Now, the $p$-group fixed-point theorem gives\[\#\op{Fix}(G/H)\equiv[G:H]\pmod p.\]The next step is to talk about what it means to be a fixed point under this action. Well, a coset $gH$ is a fixed point under multiplication by $H$ if and only if\[hgH=gH\]for any $h\in H.$ This is equivalent to $g^{-1}hgH=H,$ or $g^{-1}hg\in H.$ But this simply means that $g^{-1}Hg\subseteq H,$ and because $x\mapsto g^{-1}x\mapsto g^{-1}xg$ is a composite of bijections, we see $\#\left(g^{-1}Hg\right)=\#H,$ so in fact this is equivalent to\[g^{-1}Hg=H.\](Of course $g^{-1}Hg=H$ implies $g^{-1}Hg\subseteq H.$) So this is equivalent to $g^{-1}\in N(H)$ and $g\in N(H).$ (Here is our normalizer, as promised.)

This classifies our fixed points. Namely, our fixed points are cosets $gH$ where $g\in N(H),$ which is just the cosets in $N(H)/H.$ So the fixed-point theorem actually asserts\[\#(N(H)/H)\equiv0\pmod p.\]Because $N(H)/H$ has order divisible by $p,$ we may pick up a coset $gH$ of order $p$ by Cauchy's theorem. (We will prove Cauchy's theorem after this, but we will not pause to do so.)

It remains to construct the needed $H'.$ We take $H$ and simply extend it by $g$ to get $\langle g\rangle H$; this can be viewed set-theoretically or as a union of cosets. We get $H\subseteq H'$ for free. Further, $H'$ is a subgroup because, for any $g^ah_1,g^bh_2\in H,$ we see\[g^ah_1\cdot\left(g^bh_2\right)^{-1}\in g^aHg^{-b}=g^{a-b}H\subseteq H',\]where $g^bHg^{-b}=H$ because $g^b\in N(H).$ (This is a manifestation of $H$ being normal in $N(H).$) Lastly, $H'$ has the desired order because all the cosets $g^\bullet H$ are disjoint, and there are $p$ of them by construction of $g,$ implying $\#H'=p\cdot\#H.$ This finishes the proof. $\blacksquare$

We promised to show Cauchy's theorem, so we quickly do that now. We present a group actions proof, which aligns somewhat with our intuition.

The equivalence in the statement makes us want to use the $p$-group fixed-point theorem. The action, which comes pretty much out of nowhere, is to have $\ZZ/p\ZZ$ act on the tuples\[S:=\left\{(g_k)_{k=0}^{p-1}\in G^p:g_0\cdot\ldots\cdot g_{p-1}=e\right\}\]by cycling. Namely, we define\[k\cdot(g_0,\ldots,g_{p-1}):=(g_k,\ldots,g_{p-1+k}),\]where indices are taken$\pmod p.$ (Our intuition said that we should act by multiplication in a natural way, and this is close to that.) Now that we have this action, the proof follows quickly.

The $p$-group fixed-point theorem gives\[\#\op{Fix}(S)\equiv\#S=(\#G)^p\equiv0\pmod p.\]The next step, as before, is to talk about what a fixed point under this action is. Well, such a tuple $(g_0,\ldots,g_{p-1})$ must have, for any $k\in\ZZ/p\ZZ,$\[(g_0,\ldots,g_{p-1})=k\cdot(g_0,\ldots,g_{p-1})=(g_k,\ldots,g_{p-1+k}),\]so all entries are equal to $g_0.$ And indeed, if all entries are equal, then, $g_\ell=g_{\ell+k}$ for any $k$ or $\ell,$ implying the above holds, and we have a fixed point. Noting the condition on $S,$ we see that we have a fixed point $(g,\ldots,g)$ if and only if\[g^p=e.\]Taking a pause for a second, we see that\[\#\left\{g\in G:g^p=e\right\}\equiv0\pmod p\]from the work we've done so far. Now, a $g$ lives in the left-hand set if and only if its order divides $p,$ which means that the order is either $1$ or $p.$ Order $1$ only occurs for $g=e,$ so the remaining elements all have order $p.$ Subtracting the $g=e$ case from both sides of the above gives the desired statement. $\blacksquare$

Anyways, we now use the lemma to prove the first Sylow theorem.

We show this by induction, where the lemma is the inductive step. Intuitively, the lemma lets us start with the $p$-subgroup $H=\langle e\rangle,$ apply the given algorithm $\nu$ times, and end up with a subgroup of the needed order $p^\nu.$

To codify this, we show that we can construct a $p$-subgroup of every order $p^\alpha$ from $\alpha=0$ to $\alpha=\nu,$ by induction. For $\alpha=0,$ we choose $H=\langle e\rangle$ as our base case. For the inductive step, we may assume that we have a $p$-subgroup $H$ of order $p^\alpha$ with $\alpha+1\le\nu,$ and we want to construct $H'$ of order $p^{\alpha+1}.$ But this is exactly what the lemma does, so we are done. $\blacksquare$

While we are here, we remark that the lemma can actually show that Sylow $p$-subgroups are maximal, which is nice.

Again, we do induction, where the intuition is that the lemma lets us successively build up a Sylow $p$-subgroup from any $p$-subgroup base. Letting $\#H=p^{\nu-k},$ we prove this by induction on this $k$ as long as $k\le\nu.$ Our base case is $k=0,$ for which $H$ is immediately a Sylow $p$-subgroup.

As for the inductive step, we suppose that all $p$-subgroups of order $p^{\nu-k}$ are contained in a Sylow $p$-subgroup, and we show that all subgroups of order $p^{\nu-(k+1)}$ are contained in a Sylow $p$-subgroup, given $k+1\le\nu.$ Indeed, pick up any $H$ with $\#H=p^{\nu-(k+1)},$ and the lemma provides a $p$-subgroup $H'$ with $H\subseteq H'$ and\[\#H'=p^{\nu-(k+1)+1}=p^{\nu-k}.\]Now, $H'$ is contained in a Sylow $p$-subgroup by the inductive hypothesis, so $H$ is also. This completes the induction $\blacksquare$

With a vaguely more standard proof of the first Sylow theorem in hand, we now turn to the second and third Sylow theorems. Here is the second.

It might be tempting to let $G$ act on the Sylow $p$-subgroups by conjugation, but that will be an action helpful later. For now, we are still attached to the $p$-group fixed-point theorem, so we want a $p$-group to act.

Instead, our action will be one of the other named ones. We recall from the proof of the lemma that having a subgroup $H$ acting on cosets of $H$ made the condition $g^{-1}Hg=H$ appear as our fixed point condition, which is similar to what we want here: it shows two subgroups are conjugate (though $H$ and $H$ are conjugate trivially, of course).

Regardless, we charge ahead. Pick up two Sylow $p$-subgroups of $G$ named $P$ and $Q,$ and we will have $P$ act on the cosets $G/Q$ by multiplication. Now, $P$ is a $p$-group, so the $p$-group fixed-point theorem asserts\[\#\op{Fix}(G/Q)\equiv[G:Q]\pmod p.\]Because $[G:Q]=|G|/|Q|$ is not divisible by $p$ ($Q$ is Sylow), the above equivalence gives us a fixed point. As we saw in Cauchy's theorem, this is a more common use of the $p$-group fixed-point theorem: nonzero$\pmod p$ implies nonzero.

Anyways, we let $gQ$ be the fixed point under the action of $P.$ Arguing as we did in the lemma, this means that\[pgQ=gQ\]for any $p\in P,$ which is equivalent to $g^{-1}pg\in Q$ for any $p\in P,$ which is equivalent to $g^{-1}Pg\subseteq Q.$ But again, $x\mapsto g^{-1}xg$ is a bijection, so $g^{-1}Pg$ has the same size as $P,$ which has the same size as $Q$ because all Sylow $p$-subgroups have the same size. So in fact\[g^{-1}Pg=Q.\]It follows $P=gQg^{-1},$ so $P$ and $Q$ are indeed conjugate. This is what we wanted. $\blacksquare$

We can see the second Sylow theorem as a "power-up'' for our group actions. In particular, it is now much more meaningful to have a group or subgroup act on the group's Sylow $p$-subgroups by conjugation. Not only is this well-defined, but when the full group does it, this action is transitive by the second Sylow.

Anyways, this thinking suggests two kinds of actions we might use.

• We let a $p$-subgroup (more specifically, a Sylow $p$-subgroup) act on the Sylow $p$-subgroups by conjugations and hope to use the $p$-group fixed-point theorem.

• We have all $G$ act on the Sylow $p$-subgroups by conjugation and hope to use the transitivity.

Each of these will give a part of the third Sylow theorem. We take these in sequence.

The$\pmod p$ in the statement suggests we want to use the $p$-group fixed-point theorem, so we hope to use the first of the suggested actions. Pick up a Sylow $p$-subgroup $P$ and have it act on the set of Sylow $p$-subgroups $S_p$ by conjugation. We established in the proof of the second Sylow theorem that conjugation does not change the size of the subgroup, so this action is safe.

Now, the $p$-group fixed-point theorem gives\[\#\op{Fix}(S_p)\equiv\#S_p=n_p\pmod p.\]Thus, it remains to show that $\#\op{Fix}(S_p)\equiv1\pmod p$; in fact, we show that $\#\op{Fix}(S_p)=1.$ As usual, our next step is to examine the fixed points.

For this, we have to examine what it means for a particular Sylow $p$-subgroup $Q\in S_p$ to be a fixed point. Our action is conjugation, which means that\[pQp^{-1}=Q\]for all $p\in P.$ This is equivalent to having $p\in N(Q)$ for each $p\in P,$ or simply $P\subseteq N(Q).$ Of course, $P\subseteq N(P),$ so $P$ is one option for a fixed point, so we now have to show that $P$ is the only option.

This requires some cleverness. Note that $\#N(Q)\mid\#G$ by Lagrange's theorem, so the largest power of $p$ dividing $N(Q)$ is upper-bounded by the largest power of $p$ dividing $G.$ But $Q\subseteq N(Q)$ provides a lower bound as the largest power of $p$ dividing $G,$ so in fact $Q$ is a Sylow $p$-subgroup of $N(Q),$ and in fact so is $P\subseteq N(Q).$

The kicker is that $Q$ is normal in $N(Q),$ and by the second Sylow theorem, all Sylow $p$-subgroups are conjugate, so $P$ is conjugate to $Q,$ giving $P=Q.$ Thus, $P$ is the only fixed point of our action, and we are done. $\blacksquare$

And the other action is featured as follows.

This time around we let $G$ act on the set $S_p$ of all Sylow $p$-subgroups by conjugation. Again, this is well-defined because conjugation does not change order.

This action is transitive by the second Sylow theorem. Indeed, if $P$ and $Q$ are Sylow $p$-subgroups, then $P=gQg^{-1}$ for some $g\in G$ because $P$ and $Q$ are conjugate. This means that the orbit of $Q$ includes $P,$ giving our transitivity.

Now, we don't have access to the $p$-group fixed-point theorem because $G$ isn't a $p$-group, so we instead use the orbit-stabilizer theorem for our counting. Fix a Sylow $p$-subgroup $P,$ and we know\[\#(G\cdot P)=[G:\op{Stab}(P)].\]We just established that the action is transitive, so $\#(G\cdot P)=\#S_p=n_p.$ As for $\op{Stab}(P),$ this is the set of $g$ such that\[gPg^{-1}=P,\]which is exactly what we need for $g\in N(P).$ Thus, $\op{Stab}(P)=N(P),$ and we do indeed have $n_p=[G:N(P)].$

It remains to show that $n_p\mid m.$ Well, $P\subseteq N(P),$ so $\#P\mid\#N(P),$ and dividing from $\#G$ gives\[n_p=[G:N(P)]\mid[G:P].\]But now $[G:P]=|G|/|P|=p^\nu m/p^\nu=m.$ This gives $n_p\mid m,$ completing the proof. $\blacksquare$

We quickly remark that a lot of the above proof doesn't need to be about Sylow $p$-subgroups. If we fix $P$ as any subgroup and then let $S_p$ be the set of $P$'s conjugate subgroups, then we still get $\#S_p=[G:N(P)]$ following the same logic. We will not flesh this out here.

The above two propositions together make the third Sylow theorem, so we have proved all three Sylow theorems. I think that's enough for today.