Math Mondays

A work in progress.

Sylow Theorems

October 29, 2018

$ \newcommand{\ZZ}{\Bbb Z} \DeclareMathOperator{\Stab}{Stab} \DeclareMathOperator{\Fix}{Fix} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\sgn}{sgn} $

In group theory, the Sylow theorems are a triplet of theorems that pin down a suprising amount of information about certain subgroups.

Lagrange’s theorem tells us that if $H$ is a subgroup of $G$, then the size of $H$ divides the size of $G$. The Sylow theorems give us some answers to the converse question: for what divisors of $|G|$ can we find a subgroup of that size?


For a group $G$, and a prime $p$, and $n$ be the largest integer such that $p^n$ divides $|G|$. A $p$-subgroup of $G$ is a subgroup of order $p^k$, and if it has order $p^n$, then it is called a Sylow $p$-subgroup. Under these definitions, the Sylow theorems are:

Sylow Theorems
  1. Every $p$-subgroup is contained in a Sylow $p$-subgroup. As such, Sylow $p$-subgroups exist.
  2. All Sylow $p$-subgroups are conjugate to each other.
  3. Let $n_p$ be the number of Sylow $p$-subgroups, and $m = |G|/p^n$. Then the following hold:
    • $n_p$ divides $m$
    • $n_p \equiv 1 \bmod p$
    • $n_p = [G : N(P)]$, where $N(P)$ is the normalizer of any Sylow $p$-subgroup.

These are rather technical and deserve some more thorough digestion. Sylow 1 tells us that maximal $p$-subgroups are as big as possible; there is no obstruction preventing them from being the full $p^n$.

Sylow 2 tells us that all Sylow $p$-subgroups are isomorphic in a very strong way; there is a conjugation of the group sending them to each other. To see how this is a strong criterion, consider a non-example. Let $G = \ZZ_4 \times \ZZ_2$, and pick out the subgroups $H_1 = { (0, 0), (2, 0) }$ and $H_2 = { (0, 0), (0, 1) }$. It’s clear that $H_1$ and $H_2$ are isomorphic, but they are not conjugate. This manifests in $G/H_1 = \ZZ_2 \times \ZZ_2$ and $G/H_2 = \ZZ_4$ not being isomorphic.

Sylow 3 is the easiest to understand; it just puts some arithmetic criteria on $n_p$. For small-ish groups, this is often enough to nail down $n_p$ exactly!

On to the proofs!

Lemma

First let’s establish a lemma we’ll use frequently.

Lemma
If $G$ is a $p$-group, and it acts on a set $X$, then $\|X\| \equiv \|\Fix(X)\| \bmod p$, where $\Fix(X)$ is the set of points in $X$ that are fixed by every $g \in G$.

Proof: Let $x_1, \ldots, x_k$ be representatives for the $G$-orbits of $X$. We know that the sum of the sizes of the orbits is $|X|$. If $x_i$ is a fixed point, then the orbit is of size $1$. If it is not, then by orbit-stabilizer, the size of the orbit is $[G : \Stab(x_i)]$, which is divisible by $p$. Thus, mod $p$, every fixed point contributes $1$, and everything else in $X$ contributes $0$.

Sylow 1

Given a $p$-subgroup $H$, we show that, if it is not already maximal, we can find a $p$-subgroup $H’ \supset H$ that is $p$ times bigger. Repeating this process gives us a Sylow $p$-subgroup containing our original $H$. Since the trivial subgroup is a $p$-subgroup, this also establishes the existence of Sylow $p$-subgroups!

Let $H$ be a $p$-group that is not maximal, i.e., it has order $p^i$, where $i < n$. There is a natural action of $H$ on the left coset space $G/H$, and since $H$ is a $p$-group, our lemma tells us that $|G/H|$ is equivalent to the number of fixed points mod $p$. But since $i < n$, $G/H$ has order divisible by $p$. So the number of fixed points of this action is also divisible by $p$.

What do fixed points of this action look like? If $gH$ is a coset fixed by $h \in H$, then $hgH = gH$, i.e., $g^{-1} h g \in H$. If this is true for all $h$, then $g$ lies in the normalizer of $H$. The converse is also true, since these implications were all reversible. This means that $N(H)$ is composed of the cosets of $H$ that are fixed points.

Combining the two observations above, we conclude that $[N(H) : H]$ is divisible by $p$. Therefore, by Cauchy’s theorem, there’s some subgroup of order $p$ in $N(H)/H$. Lifting this subgroup to $N(H)$, we get a subgroup of size $p \cdot |H| = p^{i+1}$. This is the $H’$ we were looking for.

Sylow 2

Let $P$ and $Q$ be two Sylow $p$-subgroups of $G$. We want to show they are conjugate.

There is a natural action of $P$ on $G$ by multiplication, and this descends to an action of $P$ on $G/Q$ (again, left coset space). From our lemma, the number of fixed points of this action is equivalent to $|G/Q|$, mod $p$. But since $Q$ is a Sylow $p$-subgroup, $|G/Q|$ is not divisible by $p$. This means that the number of fixed points cannot be zero; i.e., there is at least one fixed point for this action. This is some $gQ$ such that $pgQ = gQ$ for all $p \in P$. Or, rearranging the terms, a $g$ such that $g^{-1}pg \in Q$ for all $p \in P$. Since $P$ and $Q$ are the same size, being Sylow $p$-subgroups, this means that $g^{-1}Pg = Q$, and so they are indeed conjugate.

Sylow 3

Let $P$ be a particular Sylow $p$-subgroup, and let it act on the set of all Sylow $p$-subgroups by conjugation. We claim that $P$ is the only fixed point of this action. This would, by our lemma (we’re getting so much mileage out of this baby), instantly tell us that $n_p \equiv 1 \bmod p$.

Consider some fixed point $Q$. Then for any $p \in P$, $p^{-1}Qp = Q$, which means that $P$ lies in the normalizer of $Q$. Since both $P$ and $Q$ are Sylow $p$-subgroups of $G$, they are both Sylow $p$-subgroups of $N(Q)$. By Sylow 2, they must be conjugate, but since $Q$ is normal in $N(Q)$, it’s not going anywhere under conjugation. Thus $Q$ must equal $P$.

Next, we show that $n_p = [G : N(P)]$. Consider the action of $G$ by conjugation on the set of Sylow $p$-subgroups. There’s only one orbit, because of Sylow 2, and by orbit-stabilizer, it has size $[G : \Stab(P)]$. But the stabilizer of $P$ is just the normalizer, so $n_p = [G : N(P)]$, as desired.

Lastly, since $m = [G : P] = [G : N(P)] [N(P) : P]$, we get that $n_p$ divides $m$ for free.

Applications

Cool! These are nice theorems, how do we put them to use? Let’s look at some example applications.


Show that $\ZZ_{35}$ is the only group of size $35$.

Let $G$ be a group of size $35$. We’ll consider its Sylow $5$ and $7$-subgroups. By Sylow 3, we know that $n_5 \equiv 1 \bmod 5$, and divides $7$. This means it’s gotta be $1$, which means $G$ has a normal subgroup of size $5$. Likewise, $n_7 \equiv 1 \bmod 7$, and divides $5$, so $G$ has a normal subgroup of size $7$ as well. They intersect trivially, since their sizes are relatively prime, so $G$ is a direct product of these groups. Therefore, $G \cong \ZZ_5 \times \ZZ_7$, which is $\ZZ_{35}$.


Classify all groups of order $105$.

Let $G$ be a group of order $105$. First, we show that it has normal Sylow $5$- and $7$-subgroups. Sylow 3 restricts $n_5 = 1,21$ and $n_7 = 1,15$.

If $n_5 = 1$, then there’s a unique Sylow $5$-subgroup $N_5$. Picking out some Sylow $7$-subgroup $P_7$, we get a subgroup $H = N_5 P_7$ of size $35$ (the normality of $N_5$ is necessary for this to be a subgroup). But from our previous exercise, we know that this must be isomorphic to $\ZZ_{35}$. Since it’s abelian, $P_7$ must of course be normal in $H$. This means that the normalizer $N(P_7) \supseteq H$. Since $n_7 = [G : N(P_7)] \le [G : H] = 3$, we are forced to conclude that $n_7 = 1$ as well.

Likewise, if $n_7 = 1$, we can construct a subgroup $H = P_5 N_7$ isomorphic to $\ZZ_{35}$, in which $P_5$ is normal. The index of $H$ here is $7$, and this also pins down $n_5 = 1$.

If neither of these are $1$, then we run out of elements. Each of these subgroups intersects trivially (because they have prime order), and so we would have $20 \cdot 4$ non-identity elements from the Sylow $5$-subgroups, and $15 \cdot 6$ non-identity elements from the Sylow $7$-subgroups. Adding in the identity, this is a total of $171$ elements, way too many.

So $G$ has normal Sylow $5$- and $7$-subgroups, and their product is a subgroup $H$ or size $35$. As the product of normal subgroups, it is itself normal. Cauchy’s theorem gives us an element $x$ of order $3$, and it generates a subgroup $K$. Since $H$ and $K$ intersect trivially, $HK$ is the whole group, and so $G$ is a semidirect product of $H$ and $K$.

What options do we have for our twisting homomorphism $\phi : K \to \Aut(H)$? All we have to do is specify $\phi(x)$, and all we need is that $\phi(x)^3$ is the identity.

The automorphisms of $\ZZ_n$ are those given by multiplying by some $a$ relatively prime to $n$. As such, the automorphisms of $\ZZ_{35}$ with degree dividing $3$ are $(r \mapsto ar)$, where $a^3 \equiv 1 \bmod 35$. The only such solutions are $1, 11, 16$.

If $a = 1$, then this is the trivial automorphism, and so $G \cong \ZZ_3 \times \ZZ_{35} \cong \ZZ_{105}$.

It turns out that the groups for $a = 11$ and $a = 16$ are isomorphic, but I can’t figure out a clean way to show it at the moment. Stay tuned.


Show $A_5$ is the smallest non-abelian simple group.

To prove this, we need to eliminate the possibility of a simple non-abelian group of any smaller size. First, we can eliminate primes; any group of size $p$ is cyclic, hence abelian.

We can also eliminate prime powers. Any group of prime power order has a non-trivial center, so it cannot be simple.

Next, we eliminate anything that is $2$ mod $4$. Such a number is equal to $2m$ with $m$ odd. If $G$ is a group of size $2m$, let $G$ act on itself by multiplication. This gives us a map $\phi : G \to S_{2m}$ sending $g$ to the permutation it induces. By Cauchy’s theorem, there’s an element of order $2$. This induces a product of $m$ transpositions, and thus an odd permutation. So the map $\sgn \circ \phi : G \to { \pm 1 }$ is surjective, and so its kernel is a non-trivial proper subgroup of $G$. (Unless $G$ has order $2$, but we already handled that case.)

Our last big sweep will be to eliminate groups of size $p^k m$ with $m < p$. Since $n_p$ divides $m$, we have $n_p \le m < p$. But $n_p$ is $1$ mod $p$, and so must be $1$. If there is a single Sylow $p$-subgroup, it must be normal. This eliminates 15, 20, 21, 28, 33, 35, 39, 44, 51, 52, 55, and 57.

This leaves us with 12, 24, 36, 40, 45, 48, and 56.

$|G|=40$: From the congruence conditions, we know that $n_5$ is $1$ mod $5$ and divides $8$. But this forces it to be $1$, so there is a unique Sylow $5$-subgroup.

$|G|=45$: Similar to $|G|=40$, the arithmetic restrictions force $n_5$ to be $1$.

$|G| = 12$: We know that $n_3$ is either $1$ or $4$. If it’s not $1$, there’s $4$ Sylow $3$-subgroups, and because they have prime order, they intersect trivially. This gives $8$ elements of order $3$, leaving $4$ other elements to constitute the Sylow $2$-subgroups. But each Sylow $2$-subgroup has $4$ elements, and so there is a unique (hence normal) one.

$|G| = 56$: Similar to the case for $12$. If $n_7$ is not $1$, it is $8$, yielding $48$ elements of order $7$. The leftover $8$ elements form the unique Sylow $2$-subgroup.

For the other three cases we need some stronger stuff.

Claim: if $G$ is simple and non-abelian, then for all $p$ dividing $|G|$, we must have $|G|$ divides $n_p!$.

Proof: Let $G$ act on the Sylow $p$-subgroups by conjugation. Because there are $n_p$ of them, this gives us a homomorphism $\phi : G \to S_{n_p}$. Since $G$ is simple, $\ker \phi$ is either trivial or all of $G$. Because all Sylow $p$-subgroups are conjugate, the latter situation only occurs when there is only one of them, something impossible if $G$ is simple and non-abelian.

This leaves us with the former case, where the kernel is trivial, and thus $\phi$ is an injection. Identifying $G$ as a subgroup of $S_{n_p}$, we get that $|G|$ divides $n_p!$ as promised.

We can now eliminate the last cases.

$|G|=24$: We know that $n_2$ is either $1$ or $3$, by the usual congruence conditions. But now we have a new tool. If $G$ were simple, then $24$ would divide $n_2!$, which it can’t in either case. So $G$ can’t be simple.

$|G|=36$: We know $n_3$ is $1$ or $4$. If $G$ is simple, then $36$ would divide $n_3!$, which it can’t.

$|G|=48$: Identical to the case for $24$.

Phew!

This was a lot of work. Back when I was in high school, we had to prove this without the Sylow theorems, and by god we appreciated them. Get off my lawn!

(But actually though, that was an… experience.)