This is the fourth post of the Introduction to Abstract Algebra series. In the first three posts, we are acquainted with the fundamental concepts and techniques in group theory. In this post, we will introduce the Sylow theorems.

The Sylow’s Theorems

Proposition 4.1: The Abelian Version of Cauchy’s Theorem

If a group $G$ is abelian and $p \mid |G|$ where $p$ is a prime, $G$ has an element of order $p$.

$Proof$. The proposition will serve as a preparation for the Sylow’s 1st theorem. We will prove this proposition by induction on the order of $G$ . Assume the proposition is true for $|G| \leq k$. For $|G|>k$, consider $x \in G, x \neq e$, and $N = <x>$. There are two cases:

• If $p \mid |x|, x^{|x|/p}$ has order $p$.
• If $p \nmid |x|$, note that $N \trianglelefteq G$ since $G$ is abelian, and consider $G/N$. Since $p \nmid |x|, p \mid |G/N|$. By the inductive hypothesis, we have an element $gN \in G/N$ such that $|gN| = p$. Note that $(gN)^{|g|} = N$, so $p \mid |g|$. Thus, $g^{|g|/p}$ is an element of order $p$.

Theorem 4.2: The Sylow’s 1st Theorem

If $G$ is a group $|G| = p^am$ where $a$ is a positive integer, $p$ is a prime and $p \nmid m$, $G$ has a subgroup of order $p^a$. We call a subgroup of order $p^a$ a Sylow p-group.

$Proof$. We will prove the Sylow’s 1st theorem by induction on the size of $G$. Assume $|G| = p^am$, consider the center $Z(G) \trianglelefteq G$, the quotient group $G/Z(G)$, and the induced homomorphism $\pi: G \to G/Z(G)$. Note that by tthe fourth isomorphism theorem, if $H < G/Z(G)$, then $\pi^{-1}(H)$ is a subgroup of $G$ of order $|H||Z(G)|$. There are two cases of the order of $G/Z(G)$:

• Case1: $p \mid |Z(G)|$. By Proposition 4.1, $Z(G)$ has a cyclic group $K$ of order $p$. Note that $K \trianglelefteq G$ and $|G/K = p^{a-1}m$. By the inductive hypothesis, $G/K$ has a Sylow p-group A of order $p^{a-1}$. Consider the induced homomorphism $\tilde \pi: G \to G/K$. Then $\tilde \pi ^{-1}(A)$ has order $p^{a}$.
• Case2: $p \nmid |Z(G)|$. Recall the class equation, $|G| = |Z(G)|+\sum_{g \in G}[G:C_G(g)]$. Since $p \nmid |Z(G)|$, $p \nmid [G:C_G(g_i)] = G/C_G(g_i)$ for some $g_i \in G$, so $p^a \mid C_G(g_i)$. Since $C_G(g_i) < G$, by the inductive hypothesis, $C_G(g_i)$ has a Sylow p-group of order $p^a$.

Theorem/Corollary 4.3: The Cauchy’s Theorem

If $G$ is a group and order of $G$ is divisible by a prime $p$, then $G$ has an element of order $p$.

$Proof$. We can prove the Cauchy’s theorem by using the Sylow’s 1st theorem. If $G$ is a group and order of $G$ is divisible by a prime $p$, then $G$ has a subgroup $H$ of order $p^m$ for some integer $m$. By the Lagarange’s theorem, $\forall x \in H, |x| = p^k$ where $k$ is an integer and $0 \leq k \leq m$. Consider the element $x^{p^{k-1}}$ in the subgroup generated by $x$, $\langle x \rangle$, $|x^{p^{k-1}}| = gcd(p^k, p^{k-1}) = p$.

Definiton 4.4: p-groups

If $G$ is a group $|G| = p^am$ where $a$ is a postive integer, $p$ is a prime and $p \nmid m$, we call a subgroup of order that is a power of $p$ a p-group, a subgroup of order $p^a$ a Sylow p-group. We denote the set of Sylow p-groups $Syl_p(G)$, the number of Sylow p-groups in $G$ as $n_p = |Syl_p(G)|$.

Theorem 4.5: The Sylow’s 2nd Theorem

If $G$ is a group, $H < G$ is a p-group, $P<G$ is a Sylow p-group, there exists a $g \in G$ such that $gHg^{-1} \leq P$.

Corollary 4.6

1. Any two Sylow p-groups are conjugate and isomorphic to each other.
2. If $P \leq G$ is a Sylow p-group, $P \trianglelefteq G$ iff $n_p = 1$.
3. If $P \leq G$ is a Sylow p-group, $n_p = [G: N_G(P)]$.

$Proof$. For the first claim, apply the Sylow’s 2nd theorem directly for $H$ being a Sylow p-group. For the second claim, both directions can be shown with the first claim and the definition of normality. For the third claim, note that the action of $G$ on $Syl_p(G$ by conjugation is transitive due to the first claim. Therefore, $n_p = |Syl_p(G)$ is the index of the stablizer of a Sylow p-group in $G$, which is $N_G(P)$.

Theorem 4.7: The Sylow’s 3rd Theorem

If $G$ is a group $|G| = p^am$ where $a$ is a positive integer, $p$ is a prime and $p \nmid m$, then $n_p \equiv 1 \mod p$ and $n_p \mid m$.

To prove the Sylow’s 2nd and 3rd theorems, we first prove a lemma.

Lemma 4.8

If $G$ is a group, $H < G$ a subgroup, $P < G$ a Sylow p-group, then $N_G(P) \cap H = P \cap H$.

$Proof$. Put $C := N_G(P) \cap H$. Clearly, $P \cap H \leq C$. We want to show that: $C \leq P \cap H \iff C \leq P \iff CP \leq P$. Since $C \leq N_G(P)$, by the second isomorphic theorem, $CP$ is a subgroup and $|CP| = |C||P|/|C\cap P|. Note that since$H$and$P$are p-groups,$p \mid |C|, p \mid |P|, p \mid |C\cap P|$, so$p \mid |CP|$. Note$P \leq CP$and$P$is a Sylow p-group, so$CP=P$. This finishes the proof.

Now we can prove the Sylow’s 2nd and 3rd theorems.

$Proof$. Suppose $Syl_p(G) = \{P_1, P_2, \cdots, P_{n_p}\}$. Let the set $S$ be the subset of $Syl_p(G)$ such that $P \in S$ is conjugate to $P_1$. We will prove the Sylow’s 2nd and 3rd theorems in four steps: (a) $|S| \equiv 1 \mod p$; (b) for all p-group $H$, $H \in P_i$ for some $P_i \in S$; (c) $|S| = n_p$; (d) $n_p \mid m$.

(a) Consider the conjugation action of $P_1$ on $S$. Suppose the orbits of the action are $O_1, O_2, \cdots, O_a$, where $O_1 = \{P_1\}$. The class equation gives

$|S| = \sum_{i=1}^a |O_i| = 1 + \sum_{i=2}^a |O_i| = 1 + \sum_{i=2}^a |P_1/stab(P_i)| = 1 + \sum_{i=2}^a |P_1/N_G(P_i) \cap P_1| = 1 + \sum_{i=2}^a |P_1/P_i \cap P_1|$

Thus, $|S| \equiv 1 \mod p$, since $p \mid |P_1/P_i \cap P_1|$.

(b) Assume for contradiction that $H$ is not contained in any $P_i \in S$. Consider the action of $H$ on $S$ by conjugation, $orb(P_i) = |H|/stab(P_i)$. Note that $stab(P_i) = N_G(P_i) \cap H = P_i \cap H$, and since $H \not\subset P_i, stab(P_i) \subset H$. Thus, $|orb(P_i)| \neq 1$, but $orb(P_i) \subseteq H$, so $orb(P_i) \mid p$. Thus, $|S| \mid p$, a contradiction to (a).

(c) This follows directly from (b), since a Sylow p-group is also a p-group. $Syl_p(G) = S$.

(d) Finally, consider the conjugation action of $G$ on $Syl_p(G)$. Note that the action is transitive. We have the class equation $n_p = |orb(P_1)| = |G|/|stab(P_1)| = |G|/|N_G(P_1)|$. Since $P_1 \subset N_G(P_1)$, $p^a \mid |N_G(P_1)| \implies n_p \mid m$.

References

[1] Siegel, Kyler. MATH GU4041 Introduction to Modern Algebra I. Department of Mathematics, Columbia University. 2019.

[2] Dummit, David and Foote, Richard. Abstract Algebra. Third Edition. John Wiley and Sons, Inc. 2004.