Sylow's Theorems

⚡ Key Takeaways

Sylow's Theorems are three structural results about finite groups asserting the existence, conjugacy, and counting of subgroups whose order is the highest power of a prime dividing the group's order. One line: if $p^n$ is the largest power of the prime $p$ dividing $|G|$ , then $G$ has a subgroup of order $p^n$ , all such are conjugate, and their number is $\equiv 1 \pmod p$ . They apply to every finite group and are the cornerstone of finite group classification. Proved by Ludwig Sylow in 1872.

💡 The Core Insight

Decompose the order of a finite group prime-by-prime. For each prime $p$ , the "biggest $p$ -piece" of $G$ — a subgroup of order equal to the full power of $p$ in $|G|$ — always exists, is unique up to conjugation, and comes in a number tightly pinned down by congruence and divisibility conditions. Group actions (especially $G$ acting on cosets and on the set of subgroups by conjugation) convert these existence/counting claims into orbit–stabilizer arithmetic.

DefinitionSylow p-subgroup

Let $G$ be a finite group of order

1

\quad |G| = p^n m, \qquad p \nmid m, \quad n \geq 1,

Derivation hint: write the prime factorization of $|G|$ and isolate the full power of the prime $p$ ; $m$ collects all other prime factors so $\gcd(p,m)=1$ . where $p$ is a prime. A Sylow $p$ -subgroup of $G$ is a subgroup $P \leq G$ with

2

\quad |P| = p^n.

Derivation hint: $p^n$ is the maximal power of $p$ dividing $|G|$ by [1]; a Sylow $p$ -subgroup attains this maximum, so it is a maximal $p$ -subgroup of $G$ . We write $n_p$ for the number of Sylow $p$ -subgroups of $G$ , and $\mathrm{Syl}_p(G)$ for the set of all of them.

TheoremSylow's Three Theorems

Let $G$ be a finite group with $|G| = p^n m$ , $p \nmid m$ .

(Sylow I — Existence.) $G$ contains at least one Sylow $p$ -subgroup; moreover, for every $k$ with $0 \le k \le n$ , $G$ has a subgroup of order $p^k$ , and every subgroup of order $p^k$ with $k<n$ is contained in one of order $p^{k+1}$ .

(Sylow II — Conjugacy.) Any two Sylow $p$ -subgroups $P, Q$ are conjugate: there is $g \in G$ with $Q = gPg^{-1}$ . Consequently every $p$ -subgroup of $G$ is contained in some Sylow $p$ -subgroup.

(Sylow III — Counting.) The number $n_p$ of Sylow $p$ -subgroups satisfies

3

\quad n_p \equiv 1 \pmod{p} \qquad \text{and} \qquad n_p \mid m.

Derivation hint: $n_p \equiv 1$ comes from $P$ acting by conjugation on $\mathrm{Syl}_p(G)$ with the single fixed point $P$ ; $n_p \mid m$ comes from $G$ acting transitively, giving $n_p = [G:N_G(P)]$ , a divisor of $[G:P]=m$ . Furthermore $n_p = [G : N_G(P)]$ , where $N_G(P)$ is the normalizer of any Sylow $p$ -subgroup $P$ .

ProofSylow I via group action on subsets

We prove existence of a subgroup of order $p^n$ .

Step 1 — Set up the action. Let $\Omega$ be the collection of all subsets of $G$ of size $p^n$ . Then

4

\quad |\Omega| = \binom{p^n m}{p^n}.

Derivation hint: choose $p^n$ elements out of the $|G|=p^n m$ elements of $G$ ; this is the binomial coefficient.

Step 2 — A $p$ -adic count. A classical lemma (Kummer/Lucas) gives that $p \nmid \binom{p^n m}{p^n}$ ; precisely the power of $p$ dividing $\binom{p^n m}{p^n}$ equals the power dividing $m$ , which is $0$ . Hence

5

\quad p \nmid |\Omega|.

Derivation hint: expand $\binom{p^n m}{p^n} = \prod_{j=0}^{p^n-1}\frac{p^n m - j}{p^n - j}$ ; for each $j$ the power of $p$ dividing $p^n m - j$ equals that dividing $p^n - j$ , so the $p$ -contributions cancel termwise.

Step 3 — Act by left multiplication. Let $G$ act on $\Omega$ by $g\cdot S = gS$ . Since $p \nmid |\Omega|$ , some orbit $\mathcal{O}$ has size not divisible by $p$ . Pick $S \in \mathcal{O}$ and let $P = \mathrm{Stab}_G(S) = \{g : gS = S\}$ .

Step 4 — Orbit–stabilizer.

6

\quad |\mathcal{O}| = [G : P] = \frac{|G|}{|P|} = \frac{p^n m}{|P|}.

Derivation hint: orbit–stabilizer theorem states the orbit size equals the index of the stabilizer; substitute $|G|=p^nm$ . Since $p \nmid |\mathcal{O}|$ , the full $p^n$ must divide $|P|$ , giving $p^n \mid |P|$ .

Step 5 — Bound $|P|$ from above. Fix $s \in S$ . The map $g \mapsto gs$ embeds $P$ into $S$ (because $PS = S$ implies $Ps \subseteq S$ ), so $|P| \le |S| = p^n$ . Combined with $p^n \mid |P|$ we get

7

\quad |P| = p^n.

Derivation hint: $P \le |S| = p^n$ from the injection $g\mapsto gs$ , while $p^n \mid |P|$ from Step 4; the only possibility is equality. Thus $P$ is a Sylow $p$ -subgroup. The chain of subgroups of orders $p^k$ follows from an inductive argument inside any $p$ -group using its nontrivial center. $\blacksquare$

∎

ProofSylow II and III via conjugation action

Let $P$ be a fixed Sylow $p$ -subgroup (exists by Sylow I), and let $X = \mathrm{Syl}_p(G)$ be the set obtained as the $G$ -orbit of $P$ under conjugation. We will show $X$ is all Sylow $p$ -subgroups.

Step 1 — $P$ acts on $X$ . Restrict the conjugation action of $G$ to the subgroup $P$ . The fixed points are those $Q \in X$ with $gQg^{-1}=Q$ for all $g\in P$ , i.e. $P \le N_G(Q)$ .

Step 2 — Fixed points are just $P$ . If $P \le N_G(Q)$ , then both $P$ and $Q$ are Sylow $p$ -subgroups of $N_G(Q)$ , and $Q \trianglelefteq N_G(Q)$ . A short argument ( $PQ$ is a $p$ -subgroup of order $|P||Q|/|P\cap Q|$ , forcing $P=Q$ ) shows the only fixed point is $P$ itself.

Step 3 — Count modulo $p$ . Since $P$ is a $p$ -group, every orbit on $X$ has size a power of $p$ ; non-fixed orbits have size divisible by $p$ . With exactly one fixed point,

8

\quad |X| \equiv 1 \pmod{p}.

Derivation hint: partition $X$ into $P$ -orbits; orbits of size $>1$ contribute multiples of $p$ , the single fixed point contributes $1$ .

Step 4 — $X$ is everything (conjugacy). Suppose some Sylow $p$ -subgroup $R \notin X$ . Let $R$ act on $X$ by conjugation; by the same argument $R$ has no fixed point in $X$ (a fixed point would force $R$ to equal a member of $X$ ). Then all $R$ -orbits have size divisible by $p$ , so $|X| \equiv 0 \pmod p$ , contradicting [8]. Hence every Sylow $p$ -subgroup lies in $X$ , i.e. all are conjugate.

Step 5 — The divisibility. By orbit–stabilizer for $G$ acting transitively on $X$ ,

9

\quad n_p = |X| = [G : N_G(P)] = \frac{|G|}{|N_G(P)|}.

Derivation hint: $X$ is a single $G$ -orbit; the stabilizer of $P$ under conjugation is exactly $N_G(P)$ ; apply orbit–stabilizer. Since $P \le N_G(P)$ , we have $p^n \mid |N_G(P)|$ , so $n_p = [G:N_G(P)]$ divides $[G:P] = m$ . Combined with [8] this gives both parts of [3]. $\blacksquare$

∎

mathematician · Peter Ludwig Mejdell Sylow | 1832–1918

A Norwegian secondary-school teacher in Halden (Frederikshald), Sylow published his three theorems in 1872 in Mathematische Annalen under the title "Théorèmes sur les groupes de substitutions," working in the language of permutation groups before the abstract group concept was standard. He spent much of his career teaching at the gymnasium level, and only late in life (1898) obtained a university chair at Christiania (Oslo), arranged in part by Sophus Lie. Together with Lie he edited the collected works of Niels Henrik Abel. His theorems are widely regarded as the deepest elementary results in finite group theory.

🧠 Intuition

Think of $|G| = p^n m$ as a number split into its $p$ -part ( $p^n$ ) and its co- $p$ -part ( $m$ ). Sylow I says the $p$ -part is always realized by an actual subgroup — the group cannot "lose" its prime-power factor; it must be packaged as a genuine subgroup. Sylow II says these prime-power packages are interchangeable: $G$ shuffles them around by conjugation, so structurally there is only "one" Sylow $p$ -subgroup up to relabeling. Sylow III is the bookkeeping: the count $n_p$ must be $1$ plus a multiple of $p$ (so $n_p \in \{1, p+1, 2p+1, \dots\}$ ) AND must divide the co- $p$ -part $m$ . Often these two constraints together leave only $n_p = 1$ , which means the Sylow subgroup is normal — an enormously useful conclusion for proving a group is not simple.

Figure

Conjugation action of a Sylow p-subgroup P on the set of all Sylow p-subgroups, illustrating why n_p ≡ 1 (mod p)

```svg

```

ExampleGroups of order 15 are cyclic

Given: $|G| = 15 = 3 \cdot 5$ . Take $p = 5$ , so here $p^n = 5$ , $m = 3$ ; and $p=3$ , $p^n=3$ , $m=5$ .

Step 1 — Count Sylow 5-subgroups. $n_5 \equiv 1 \pmod 5$ and $n_5 \mid 3$ . Divisors of $3$ are $\{1,3\}$ ; only $1 \equiv 1 \pmod 5$ . So $n_5 = 1$ .

Step 2 — Count Sylow 3-subgroups. $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 5$ . Divisors of $5$ are $\{1,5\}$ ; only $1 \equiv 1 \pmod 3$ (since $5 \equiv 2$ ). So $n_3 = 1$ .

Step 3 — Both are normal. $n_5 = 1$ means the unique Sylow 5-subgroup $P_5 \trianglelefteq G$ ; likewise $P_3 \trianglelefteq G$ . They intersect trivially ( $\gcd(3,5)=1$ ), and $|P_3 P_5| = 15 = |G|$ .

Step 4 — Internal direct product. $G \cong P_3 \times P_5 \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}$ .

\boxed{\text{Every group of order } 15 \text{ is cyclic, isomorphic to } \mathbb{Z}_{15}.}

ExampleNo simple group of order 30

Given: $|G| = 30 = 2 \cdot 3 \cdot 5$ .

Step 1 — Sylow 5. $n_5 \equiv 1 \pmod 5$ , $n_5 \mid 6$ . Candidates dividing 6: $\{1,2,3,6\}$ ; those $\equiv 1 \pmod 5$ : $\{1,6\}$ . So $n_5 \in \{1,6\}$ .

Step 2 — Sylow 3. $n_3 \equiv 1 \pmod 3$ , $n_3 \mid 10$ . Candidates: $\{1,2,5,10\}$ ; those $\equiv 1 \pmod 3$ : $\{1,10\}$ . So $n_3 \in \{1,10\}$ .

Step 3 — Suppose neither is normal. Then $n_5 = 6$ and $n_3 = 10$ . Distinct Sylow 5-subgroups have order 5 and meet only in the identity, so they contribute $6\cdot(5-1) = 24$ elements of order 5. Distinct Sylow 3-subgroups contribute $10\cdot(3-1) = 20$ elements of order 3.

Step 4 — Count overflow. That is $24 + 20 = 44$ non-identity elements, but $|G| = 30$ . Contradiction.

Step 5 — Conclusion. Hence $n_5 = 1$ or $n_3 = 1$ ; a unique Sylow subgroup is normal.

\boxed{\text{No group of order } 30 \text{ is simple — it has a nontrivial normal subgroup.}}

⚠️ Common mistake

$n_p \mid m$ , not $n_p \mid |G|$ in general — though $n_p \mid |G|$ is also true but weaker. The sharp divisibility is $n_p \mid m$ , the co- $p$ -part. Using only $n_p \mid |G|$ throws away information. Also, $n_p \equiv 1 \pmod p$ is a separate condition from $n_p \mid m$ — you must apply both. Another frequent error: assuming a Sylow $p$ -subgroup is unique automatically; uniqueness ( $n_p=1$ ) is precisely the special case where it is normal, and must be deduced, not assumed.

⚠️ Common mistake

Sylow's theorems do NOT give subgroups of every order dividing $|G|$ . They guarantee subgroups of prime-power orders $p^k$ (for $k \le n$ ), not, say, a subgroup of order $m$ or of an arbitrary divisor. The general converse of Lagrange is false.

CounterexampleLagrange has no full converse

The alternating group $A_4$ has order $12 = 2^2 \cdot 3$ . Although $6 \mid 12$ , $A_4$ has no subgroup of order 6. (Any such subgroup would have index 2, hence be normal and contain all squares of elements; counting the eight 3-cycles forces a contradiction.) Sylow's theorems correctly predict only subgroups of orders $1,2,3,4$ — and indeed $A_4$ has a normal Sylow 2-subgroup $V_4$ of order 4 ( $n_2 = 1$ ) and four Sylow 3-subgroups ( $n_3 = 4 \equiv 1 \pmod 3$ , $4 \mid 3$ ? no — recheck: $m=3$ , $n_3 \mid 3$ and $\equiv 1 \pmod 3$ gives $n_3 \in\{1\}$ ... but $A_4$ actually has $n_3=4$ ). The resolution: for $p=3$ , $|A_4|=2^2\cdot 3$ so $m=4$ , $n_3 \mid 4$ , $n_3 \equiv 1 \pmod 3 \Rightarrow n_3\in\{1,4\}$ , and indeed $n_3 = 4$ . This shows the theorems are tight but the order-6 subgroup simply does not exist.

connection

Sylow's theorems sit at the heart of finite group theory and link to many neighbors. They refine Lagrange's Theorem by providing a partial converse. Their proofs are pure applications of Group Actions and the Orbit-Stabilizer Theorem, and the counting argument is the Class Equation specialized to conjugation on subgroups. They feed directly into the Classification of Finite Simple Groups by ruling out simplicity for many orders, and the local-global philosophy they embody (studying $G$ one prime at a time) generalizes into Local Analysis in Group Theory and the theory of p-Groups and Nilpotent Groups (a finite group is nilpotent iff it is the direct product of its Sylow subgroups).

🔬 Where this appears

⏳ History

- 1770–1771 — Lagrange. Joseph-Louis Lagrange studies permutations of roots of equations, implicitly the "order divides" idea later named Lagrange's theorem.

- 1801 — Gauss. Disquisitiones Arithmeticae contains group-theoretic reasoning on residues and forms, prefiguring abelian group structure.

- 1830s — Galois. Évariste Galois introduces normal subgroups and the role of group structure in solvability of equations.

- 1845 — Cauchy. Augustin-Louis Cauchy proves that if a prime

p

divides

|G|

then

G

has an element (hence subgroup) of order

p

— the existence seed Sylow would vastly generalize.

- 1872 — Sylow. Ludwig Sylow publishes the three theorems for permutation groups in Mathematische Annalen, the definitive existence/conjugacy/counting results.

- 1878 — Frobenius. Georg Frobenius gives a proof valid for abstract (not merely permutation) groups, freeing the theorems from the substitution-group formalism.

- 1887 — Hölder & others. Sylow's theorems become central tools in the program to classify groups by order; Hölder classifies groups of several small orders.

- 1895–1911 — Burnside. William Burnside's Theory of Groups of Finite Order canonizes Sylow theory and uses it (with character theory) toward results like the

p^a q^b

solvability theorem.

- 1960s–1980s — Classification. Sylow analysis is a recurrent engine in the decades-long classification of finite simple groups (Feit–Thompson, Gorenstein, Aschbacher, et al.).

open-problem

While Sylow's theorems are fully proved and complete, deep questions about Sylow structure remain active. The McKay conjecture (recently announced as proved, building on the work of many) asserts a numerical coincidence between irreducible characters of $G$ of degree prime to $p$ and those of the normalizer $N_G(P)$ of a Sylow $p$ -subgroup — a profound "local determines global" refinement. The broader local–global principle (Alperin's weight conjecture, the Brauer height-zero conjecture) seeks to express invariants of $G$ entirely in terms of $p$ -local subgroups built from Sylow subgroups; several of these remain open or only recently settled. Determining the precise structure (not just existence and count) of Sylow subgroups in arbitrary families of groups, and how the integer $n_p$ behaves, is genuinely subtle: e.g., which integers $n$ can occur as $n_p$ for some group is itself a studied question.

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also