DEC 12, 2024

an attempt to understand the sylow theorems

The following are some notes I wrote in an attempt to understand the Sylow theorems and their proofs.^[1] It draws heavily from Keith Conrad’s expository notes and T. W. Hungerford’s Algebra (1974), as well as the presentation of the material in MATH 230A at UCI (taught by Professor Vladimir Baranovsky). I have also referenced some course materials and videos by Nathan Kaplan and occasionally, Dummit and Foote’s Abstract Algebra (3rd edition).

Use at your own risk! I promise that I haven’t intentionally included errors, but humans will be human. Please feel free to email me if you find any mistakes.

why care about the sylow theorems?

Lagrange’s Theorem states that any subgroup of a finite group $G$ must have order $d$ dividing $|G|$. After learning of this fact, you might wonder whether the converse is true. If $d \mid |G|$, does $G$ have a subgroup of order $d$?

You might do some preliminary exploration. If you know the fundamental theorem of finite abelian groups, you would conclude that the converse does indeed hold for finite abelian groups. Afterward, you might try some small examples like $S_6$ and $D_4$, which do in fact have a subgroup of order $d$ for every $d$ dividing their order. You might start to hope—maybe the converse is true in general?

Alas, if you kept looking, you would quickly find a counterexample. The smallest (and thus most commonly referenced) counterexample is the group $A_4$, which, despite having order $12$, does not have any subgroups of order $6$.^[2] So then the question that arises is this: when is it true that $d\mid |G|$ implies that $G$ has a subgroup of order $d$? And what can we say about these subgroups?

A starting point is Cauchy’s theorem, which states that there is a subgroup of order $d$ if $d$ is prime. We can use this theorem to prove something a more powerful—the first Sylow theorem, which says that there is always a subgroup of order $p^i$ for $p^i$ dividing $|G|$. These subgroups of prime power order are called $p$-subgroups. The second and third Sylow tell us a bit more about the biggest of these subgroups—subgroups of order $p^k$, where $p^k$ is the largest power of $p$ that divides $|G|$. We call these subgroups of maximal prime power order Sylow $p$-subgroups. In short, Sylow II states all Sylow $p$-subgroups are conjugate to each other, and Sylow III states that the number of distinct Sylow $p$-subgroups is congruent to $1 \pmod p$ and divides $|G|$.

a note on the proofs

In the following, I assume that you are familiar with the definition and basic properties of group actions, the orbit-stabilizer theorem, and the normalizer of a subgroup.

We’ll cover the proofs of Cauchy’s theorem and the three Sylow theorems. All of these proofs go something like this: look at a particular group action, where the acting group is of prime power order for some prime $p$. Then consider the number of fixed points of that action $\bmod p$, using a lemma called the fixed point congruence.^[3]

the fixed point congruence

As promised, let’s look at the number of fixed points of a group action. The particular statement that will be useful to us is the following.

Lemma. (Fixed point congruence). Let $P$ be a group of order $p^i$ for some prime $p$, and let $P$ act on a set $X$. Then
$$|X| \equiv |\text{fix}_P(X)| \pmod p$$
where $\text{fix}_P(X) = \set{x\in X \mid gx = x \text{ for all } g \in P}$.

Idea. We count the sizes of the orbits of fixed points separately from the sizes of the orbits of non-fixed points. Then, we rewrite the sizes of the orbits of non-fixed points using the orbit-stabilizer theorem, and reduce $\bmod p$.

Proof. We know that the orbits in $X$ partition $X$, so we can compute $|X|$ by adding up the sizes of the orbits. Since the orbit of a fixed point has exactly one element, the total size of these orbits is exactly $|\text{fix}_P(X)|$. For the orbits that have more than one element, we can pick a representative from each, say $x_1,…, x_n$, and sum up the size of each orbit. Then we end up with

$$|X| = |\text{fix}_P(X)| + \sum_{i=1}^{n} |\text{orb}(x_i)|.$$

By the orbit-stabilizer theorem, we know that $|\text{orb}(x)|=[P:\text{stab}(x)]$ for any $x \in X$. But $P$ has order $p^k$, so as long as $\text{stab}(x)$ is not $P$ itself (that is, as long as $x$ is not a fixed point), we have that $p\mid [P:\text{stab}(x)]$. So we know $p \mid |\text{orb}(x_i)|$ for all $i$ in our summation above. Reducing $\bmod p$, we conclude that
$$|X| \equiv |\text{fix}_P(X)| \pmod p$$
which is what we wanted.

cauchy’s theorem

Theorem. (Cauchy). If a prime $p$ divides the order of a finite group $G$, $G$ has a (cyclic) subgroup of order $p$.

Idea. We construct a special set of $p$-tuples, then have $\mathbb{Z}_p$ act on this set by “shifting over” the elements of the tuple by the corresponding number of spaces. Using the fixed point congruence, we can find a nontrivial fixed point of this action. This gives us an element of order $p$, which generates a subgroup of order $p$.

Proof. Consider the following set of $p$-tuples of elements of $G$:
$$X=\set{(x_1,x_2,…,x_p) \mid x_1x_2\cdots x_p=e}.$$
We can see that $|X|=|G|^{p-1}$, because once we choose $x_1, x_2, …, x_{p-1}$, our only choice for $x_p$ is $x_p =(x_1x_2\cdots x_{p-1})^{-1}$. This implies that $p\mid |X|$, since $p \mid |G|$.

Now let $\mathbb{Z}_p$ act on $X$ as follows:
$$k(x_1, x_2, …, x_p)=(x_{k+1},x_{k+2}, …, x_p,x_1, …, x_k).$$
We now verify that this group action is well-defined. It is easy to see that for any $x \in X$, $0x=x$ and $(k_1+k_2)x =k_1(k_2x)$. We claim that $(x_{k+1},x_{k+2}, …, x_p,x_1, …, x_k) \in X$. Observe that since $x_1x_2\cdots x_p=(x_1\cdots x_k)(x_{k+1}\cdots x_p)=e$, we know that $(x_1\cdots x_k)^{-1} =(x_{k+1}\cdots x_p)$. So $x_{k+1}x_{k+2}\cdots x_p x_1 \cdots x_k=(x_1\cdots x_k)^{-1}(x_1\cdots x_k)=e$, which means $(x_{k+1},x_{k+2}, …, x_p,x_1, …, x_k) \in X$. Hence, the action is well-defined.

By the fixed point congruence, we know that
$$|X|\equiv |\text{fix}_{\mathbb{Z}_p}(X)|\pmod p.$$
Recalling that $p \mid |X|$, we see that $p\mid |\text{fix}_{\mathbb{Z}_p}(X)|$. Next, observe that $\text{fix}_{\mathbb{Z}_p}(X)$ consists of exactly the $p$-tuples where all $x_i$'s are the same. In particular, we can see that $(e, e, …, e) \in \text{fix}_{\mathbb{Z}_p}(X)$. So $|\text{fix}_{\mathbb{Z}_p}(X)| >0$, which means that there are at least $p$ elements in $\text{fix}_{\mathbb{Z}_p}(X)$. Then we can find an element $(a, a, …, a) \in X$ with $x \neq e$, which precisely means that $a^p=e$. Because $p$ is prime, we know that this element $a$ has order $p$, so $\langle a \rangle$ is a subgroup of $G$ with order $p$.

sylow I

Theorem. (Sylow I). Let $G$ be a group, and let $p^k$ be the highest power of a prime $p$ that divides $|G|$. Then $G$ has a subgroup of order $p^i$ for each $0\leq i\leq k$.

Idea. For Sylow I, we consider a $p$-subgroup $H$ acting on $G/H$ by left multiplication. We start by noting that the trivial group is always a subgroup of order $p^0$, then proceed by induction to generate subgroups of bigger and bigger prime power.

Proof. Clearly, $G$ has a subgroup of order $p^0=1$, since $\langle e \rangle \leq G$. Now suppose $H$ is a subgroup of $G$ with order $p^i$, where $0\leq i <k$. We will show that there exists a subgroup $H’$ of order $p^{i+1}$.

Let $H$ act on $G/H$ by left multiplication. We know that $gH \in G/H$ is a fixed point if and only if $hgH =gH$ for all $h\in H$. This is the same as saying $g^{-1}hg \in H$ for all $h \in H$. So $gH$ is a fixed point if and only if $g$ normalizes $H$. That is, $\text{fix}_H(G/H)=N(H)/H$. Then since $H$ has prime power order, we know by the fixed point congruence that
$$|G/H| \equiv |\text{fix}_H(G/H)| \pmod p,$$
which allows us to conclude that
$$|G/H| \equiv |N(H)/H| \pmod p.$$
Recall that $|G| = p^k$ and $|H| = p^i$ with $0\leq i < k$, so $p\mid |G/H|$. Then $p \mid |N(H)/H|$ by the above. Since $H$ is always normal in its normalizer, we know that $N(H)/H$ is a group. So we may apply Cauchy’s theorem to see that $N(H)/H$ has a subgroup $P$ of order $p$. We know $P=H’/H$ for some $H’ \leq N(H) \leq G$.^[4] Since $|H’/H| = p$ and $|H|=p^{i}$, $|H’|=p^{i+1}$.

By induction, we conclude that $G$ has a subgroup of order $p^i$ for each $0\leq i\leq k$.

sylow II

Theorem. (Sylow II). Any two Sylow-$p$ subgroups of $G$ are conjugate.

Idea. We consider two Sylow-$p$ subgroups of $G$, $P$ and $Q$, and let $Q$ act on $G/P$ by left multiplication. As in the proof of Sylow I, we apply the fixed point congruence, and show $P$ and $Q$ are conjugate using the fact that there is at least one fixed point of the action.

Proof. Let $P$, $Q$ be two Sylow-$p$ subgroups of $G$, and let $Q$ act on $G/P$ by left multiplication. By the fixed point congruence, we know that
$$|G/P| \equiv |\text{fix}_Q(G/P)| \pmod p.$$
Since $|P|$ is the highest power of $p$ dividing $|G|$, we know that $p\nmid|G/P|$. Then we can see that $|\text{fix}_Q(G/P)|\not\equiv0 \bmod p$, so $|\text{fix}_Q(G/P)|\neq0$. This means there is at least one fixed point of the group action, say $gP$ for some $g \in G$. Since $gP$ is a fixed point, $qgP=gP$ for all $q \in Q$. That is, $g^{-1}qgP = P$ for all $q \in Q$, which means $g^{-1}Qg=P$.

We conclude that $P$, $Q$ are conjugate, as desired.

sylow III

Theorem. (Sylow III). The number of distinct Sylow-$p$ subgroups of $G$, is congruent to $1 \bmod p$ and divides $|G|$.

Idea. Denote the set of Sylow-$p$ subgroups of $G$ by $\text{Syl}_p(G)$. To show $|\text{Syl}_p(G)| \equiv 1 \pmod p$, we consider a Sylow-$p$ subgroup $P$ acting by conjugation on $\text{Syl}_p(G)$, then apply the fixed point congruence. Since the action is conjugation, apply Sylow II to get some extra information about the set of fixed points. To show $|\text{Syl}_p(G)|$ divides $|G|$, let the whole group $G$ acting on $\text{Syl}_p(G)$ by conjugation. We conclude by using Sylow II, the first part of Sylow III, and the orbit-stabilizer theorem.

Proof. First, consider a Sylow-$p$ subgroup $P$ of $G$, which we know exists by Sylow I. Let $P$ act on $\text{Syl}_p(G)$ by conjugation. By the fixed point congruence, we have
$$|\text{Syl}_p(G)| \equiv |\text{fix}_P(\text{Syl}_p(G))| \pmod p.$$
Suppose that $Q$ is a fixed point of the group action. We claim that $Q =P$. First, observe that $P \in\text{fix}_P(\text{Syl}_p(G))$, since conjugating by elements in $P$ does not change $P$. Also, note that $Q \subseteq N(Q)$. Since conjugation by elements of $P$ fixes $Q$, $P \subseteq N(Q)$ as well. We also know that because $|N(Q)|$ divides $|G|$, $P$ and $Q$ are Sylow-$p$ subgroups in $N(Q)$. So they are conjugate by Sylow II. But $Q$ is normal in $N(Q)$, so any conjugate of $Q$ is $Q$ itself. Hence, $Q=P$, and $|\text{fix}_P(\text{Syl}_p(G))|=1$. We conclude that
$$|\text{Syl}_p(G)|\equiv 1 \pmod p.$$
Next, consider $G$ acting on $\text{Syl}_p(G)$ by conjugation. Because all the Sylow-$p$ subgroups are conjugate by Sylow II, we know that there is only one orbit in $\text{Syl}_p(G)$. That is, any orbit has size $|\text{Syl}_p(G)|$. So by the orbit-stabilizer theorem, $|\text{Syl}_p(G)|$ divides $|G|$.