# Sylow Theorems

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

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.)