TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Sylow I: every finite group has a Sylow $p$ -subgroup

Statement

Let $G$ be finite, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G$ has a subgroup of order $p^{a}$ , hence a Sylow $p$ -subgroup (Sylow $p$ -subgroups of a finite group). See Sylow $p$ -subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let $G$ be a finite group, let $p$ be prime, and write $∣ G ∣ = p^{a} m$ with $a \in N$ and $p ∤ m$ . A subgroup $P \leq G$ is a Sylow $p$ -subgroup when $∣ P ∣ = p^{a}$ . Equivalently, its order is the largest power of $p$ dividing $∣ G ∣$ . This is a property of a subgroup and does not presume that such a subgroup exists; existence is proved in thm-sylow-first-theorem. (Sylow $p$ -subgroups of a finite group).

[L2]

Let $p$ be prime and let $a, m \in N$ satisfy $p ∤ m$ . Then $v_{p} ⁣ ((\binom{p^{a} m}{p^{a}})) = 0.$ The valuation is applied only to nonzero integers. (If $∣ G ∣ = p^{a} m$ with $p ∤ m$ , then $v_{p} (\binom{p^{a} m}{p^{a}}) = 0$ ).

[L3]

Let $G$ act on $X$ and let $x \in X$ . The rule $Φ : G / G_{x} ⟶ G \cdot x, Φ (g G_{x}) = g \cdot x,$ is well-defined and bijective. Thus every orbit is naturally in bijection with the left cosets of its stabilizer. (Orbit-stabiliser: $G / G_{x} \to G \cdot x$ , $g G_{x} \mapsto g \cdot x$ , is a well-defined bijection).

[L4]

For an action of $G$ on $X$ and $x \in X$ , $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite. In particular, if $G$ is finite, then $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣ .$ . (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ).

[L5]

Let $G$ be a finite group and $H \leq G$ . Then $∣ G ∣ = [G : H] ∣ H ∣ .$ Consequently, under the canonical embedding $ι : N \to Z$ , $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

Let $P$ be a property of naturals such that for every $n \in N$ , if $P (m)$ holds for all $m < n$ then $P (n)$ . Then $P (n)$ holds for all $n \in N$ . (At $n = 0$ the hypothesis is vacuous, so $P (0)$ is forced.). (Strong (complete) induction).

[L7]

For a left action of $G$ on $X$ , the relation $x \sim y$ defined by $y = g \cdot x$ for some $g \in G$ is an equivalence relation whose class at $x$ is $G \cdot x$ , and the distinct orbits partition $X$ (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set).

Proof

technique · direct

1.1L2L6givenalgebra

Argue by strong induction [L6] on $∣ G ∣$ , the induction statement being that every finite group of order $n$ has a subgroup of order $p^{a}$ whenever $n = p^{a} m$ with $p ∤ m$ . Let $Ω$ be the set of subsets of $G$ of size $p^{a}$ and let $G$ act on $Ω$ by left translation, $g \cdot A : = g A$ ; this is an action, and $∣ g A ∣ = ∣ A ∣$ because left translation is a bijection of $G$ . Counting subsets gives $∣ Ω ∣ = (\binom{p^{a} m}{p^{a}})$ , so [L2] yields $v_{p} (∣ Ω ∣) = 0$ , that is $p ∤ ∣ Ω ∣$ .

2.1step 1.1L3L4L7choose

By [L7] the orbits partition $Ω$ , so $∣ Ω ∣$ is the sum of the orbit sizes. Were $p$ to divide every orbit size it would divide $∣ Ω ∣$ , so some orbit $O$ has $p ∤ ∣ O ∣$ . Choose $A \in O$ and put $H = G_{A} = {g \in G : g A = A}$ . The bijection of [L3] between $G / H$ and $O$ gives $∣ O ∣ = [G : H]$ , so [L4] gives $p^{a} m = ∣ G ∣ = ∣ H ∣ ∣ O ∣$ ; since $p ∤ ∣ O ∣$ , the full power $p^{a}$ divides $∣ H ∣$ .

3.1step 2.1givenalgebra

Suppose $H = G$ . Then $g A = A$ for every $g \in G$ , so for any $x \in A$ the set $A$ contains $G x = G$ , whence $A = G$ and $p^{a} = ∣ A ∣ = ∣ G ∣ = p^{a} m$ . Thus $m = 1$ and $G$ itself is a subgroup of order $p^{a}$ .

3.2step 2.1L5L6givenalgebra

Suppose instead $H \neq G$ , so $∣ H ∣ < ∣ G ∣$ . By [L5], $∣ H ∣$ divides $p^{a} m$ ; writing $∣ H ∣ = p^{b} m^{'}$ with $p ∤ m^{'}$ , step 2.1 gives $b \geq a$ , while $p^{b} ∣ p^{a} m$ with $p ∤ m$ gives $b \leq a$ . Hence $∣ H ∣ = p^{a} m^{'}$ with $p ∤ m^{'}$ , and the induction hypothesis applied to $H$ supplies a subgroup of $H$ of order $p^{a}$ , which is a subgroup of $G$ .

4.1L1step 3.1step 3.2given∎

Steps 3.1 and 3.2 are exhaustive, so $G$ has a subgroup $P$ of order $p^{a}$ , and $∣ P ∣$ is the largest power of $p$ dividing $∣ G ∣$ , so $P$ is a Sylow $p$ -subgroup by [L1]. At $a = 0$ the argument returns the trivial subgroup, of order $p^{0} = 1$ ; for the trivial group $∣ G ∣ = 1$ this is $G$ itself, and $m = 1$ is the case settled in step 3.1.

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Amin Idelhaj, The Sylow Theorems and Their Applications, Section 3, Lemma 3.6 and the proof of Sylow's first theorem (standard reference, not scraped)
Keith Conrad, The Sylow Theorems, Section 2, Proof of Sylow I (standard reference, not scraped)

Sylow I: every finite group has a Sylow pp-subgroup