The Sylow Theorems

Let $G$ be a finite group of order $|G| = p^n m$ where $p$ is prime and $\gcd(p, m) = 1$. A Sylow $p$-subgroup of $G$ is a subgroup of order $p^n$.

Theorem 7.1 (Sylow”s First Theorem). $G$ has a Sylow $p$-subgroup.

Theorem 7.2 (Sylow’s Second Theorem). Any two Sylow $p$-subgroups are conjugate.

Theorem 7.3 (Sylow’s Third Theorem). The number $n_p$ of Sylow $p$-subgroups satisfies:

1. $n_p \equiv 1 \pmod{p}$.
2. $n_p$ divides $m$.

Proof. Let $|G| = p^n m$ with $\gcd(p, m) = 1$. Let $X$ be the set of all subsets of $G$ of size $p^n$. Then $|X| = \binom{p^n m}{p^n}$. Note that $p$ does not divide $\binom{p^n m}{p^n}$ (this follows from Lucas’s theorem or examining the $p$-adic valuation). $G$ acts on $X$ by left multiplication. Since $|X|$ is not divisible by $p$Some orbit $\mathrm{Orb}(S)$ has size not divisible by $p$. By the orbit-stabilizer theorem, $|\mathrm{Stab}(S)| = |G|/|\mathrm{Orb}(S)|$ is divisible by $p^n$. For $s \in S$Left multiplication by $s$ is a bijection $S \to sS$And $sS \subseteq S$ since $\mathrm{Stab}(S) \cdot S = S$. Since $|sS| = |S| = p^n$We get $sS = S$So $s \in \mathrm{Stab}(S)$. Thus $S \subseteq \mathrm{Stab}(S)$Giving $p^n \leq |\mathrm{Stab}(S)|$. Since $|\mathrm{Stab}(S)|$ divides $p^n m$ and is divisible by $p^n$We have $|\mathrm{Stab}(S)| = p^n$ And $\mathrm{Stab}(S)$ is a Sylow $p$-subgroup. $\blacksquare$

Proposition 7.4. Every group of order $pq$ (where $p \lt q$ are primes with $p \nmid (q - 1)$) Is cyclic.

Proof. By Sylow’s third theorem, $n_q \equiv 1 \pmod{q}$ and $n_q$ divides $p$. Since $p \lt q$ $n_q = 1$So the Sylow $q$-subgroup $Q$ is normal. Similarly, $n_p \equiv 1 \pmod{p}$ and $n_p$ Divides $q$. Since $p \nmid (q-1)$, $n_p \neq q$So $n_p = 1$And the Sylow $p$-subgroup $P$ Is normal. Since $P \cap Q = \{e\}$ (their orders are coprime) and $|PQ| = pq = |G|$ We have $G \cong P \times Q \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} \cong \mathbb{Z}/pq\mathbb{Z}$. $\blacksquare$

Proposition 7.5. Every group of order $p^2$ (where $p$ is prime) is abelian.

Proof. Let $|G| = p^2$. If $G$ is cyclic, it is abelian. Otherwise, every non-identity element has Order $p$ (by Lagrange). Let $g \neq e$ and consider $H = \langle g \rangle$ with $|H| = p$. Then $[G : H] = p$So $H \trianglelefteq G$ (the smallest prime dividing $|G|$). Pick $x \notin H$. Then $G = H \cup xH$And since $x \notin H$ and $\langle x \rangle$ has order $p$ We have $G = H \times \langle x \rangle$Which is abelian. $\blacksquare$

Proof. Let $P$ be a Sylow $p$-subgroup of $G$And let $Q$ be any $p$-subgroup of $G$. $Q$ acts on the set of left cosets $G/P$ by left multiplication: $q \cdot (gP) = qgP$.

Since $|G/P| = |G|/|P| = m$ is not divisible by $p$And orbits under the $Q$-action have sizes Dividing $|Q|$ (hence powers of $p$), the number of fixed points satisfies:

$|\mathrm{Fix}(Q)| \equiv |G/P| \equiv m \not\equiv 0 \pmod{p}$

So there exists $gP \in G/P$ fixed by $Q$Meaning $Q \cdot gP = gP$I.e., $QgP = gP$So $g^{-1}Qg \subseteq P$.

Taking $Q$ to be a Sylow $p$-subgroup: $|g^{-1}Qg| = |Q| = p^n = |P|$So $g^{-1}Qg = P$ Proving that $P$ and $Q$ are conjugate. $\blacksquare$

Proof. Let $P$ be a Sylow $p$-subgroup. $P$ acts on the set $\mathrm{Syl_p}(G)$ of all Sylow $p$-subgroups by conjugation. Write $\mathrm{Syl_p}(G) = \{P = P_1, P_2, \ldots, P_{n_p}\}$.

Step 1: $n_p \equiv 1 \pmod{p}$. A Sylow $p$-subgroup $P_i$ is a fixed point of the $P$-action Iff $P \subseteq N_G(P_i)$. But then $PP_i \leq N_G(P_i)$And $|PP_i| = p^n \cdot p^n / |P \cap P_i|$ Which is a power of $p$. Since $p^n$ is the maximal power of $p$ dividing $|G|$ and $PP_i \subseteq G$ We get $|PP_i| = p^n$Hence $P = PP_i = P_i$ (since $P \subseteq PP_i$).

Thus $P$ is the unique fixed point. All other orbits have size $[P : \mathrm{Stab_P}(P_i)]$ A power of $p$ greater than $1$. By the fixed-point congruence for $p$-group actions: $n_p = 1 + \sum (\mathrm{sizes\ of\ remaining\ orbits}) \equiv 1 \pmod{p}$. Step 2: $n_p$ divides $m$. The group $G$ acts transitively on $\mathrm{Syl_p}(G)$ by conjugation (by Sylow’s second theorem). Hence $n_p = |\mathrm{Syl_p}(G)| = [G : N_G(P)]$. Since $P \leq N_G(P)$, $|N_G(P)|$ is divisible by $p^n$. Therefore $n_p = |G|/|N_G(P)|$ divides $|G|/p^n = m$. $\blacksquare$

Problem. Find all Sylow $2$-subgroups and Sylow $3$-subgroups of $S_4$.

Proposition 7.6. Every group of order $15$ is cyclic.

Proof. $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $3$So $n_5 = 1$. $n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $5$So $n_3 = 1$. Both the Sylow $3$-subgroup $P$ and the Sylow $5$-subgroup $Q$ are Normal, $P \cap Q = \{e\}$And $PQ = G$. Hence $G \cong P \times Q \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}$. $\blacksquare$

Proposition 7.7. If $G$ is a simple group with $|G| \leq 60$Then $|G|$ is prime. The smallest non-abelian simple group is $A_5$ of order $60$.

Proof sketch. If $G$ is simple and $|G| = p^n m$ with $\gcd(p, m) = 1$ and $m > 1$ Then $n_p = m$ (since $n_p \neq 1$And $n_p$ divides $m$ with $n_p \equiv 1 \pmod{p}$). For many orders, $n_p = 1$Forcing a normal Sylow subgroup and contradicting simplicity. $\blacksquare$

:::caution Common Pitfall Sylow subgroups are not unique . When $n_p > 1$There are multiple Sylow $p$-subgroups, But they are all conjugate. A common mistake is to assume $n_p = 1$ without checking the Sylow Conditions. Always verify that $n_p \equiv 1 \pmod{p}$ and $n_p$ divides $m$.

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