TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$

Statement

Let $G$ be a finite group and let $p$ be prime. If $p ∣ ∣ G ∣$ , then $G$ contains an element of order $p$ .

Facts & Assumptions

Given: A finite group $G$ and a prime $p$ dividing $∣ G ∣$ .

[L1]

A finite $p$ -group acting on a finite set satisfies the fixed-point congruence (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L2]

The additive group $Z / p$ is a group with $p$ elements and hence is a finite $p$ -group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L3]

If $A$ and $B$ are finite, the set of functions $B \to A$ has cardinality $∣ A ∣^{∣ B ∣}$ (The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $∣ A^{B} ∣ = ∣ A ∣^{∣ B ∣}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ ).

[L4]

A prime is greater than $1$ and has only $1$ and itself as positive divisors (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L5]

A congruence $a \equiv b (m o d p)$ means that $p$ divides $a - b$ , and divisibility means existence of an integer factor (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L6]

The order of an element is the least positive exponent giving the identity (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · constructive

1.1

Let $Ω$ be the set of $p$ -tuples $(g_{0}, \dots, g_{p - 1}) \in G^{p}$ whose ordered product is $e$ . The first $p - 1$ coordinates determine the last uniquely as $(g_{0} \dots g_{p - 2})^{- 1}$ , so [L3] gives $∣ Ω ∣ = ∣ G ∣^{p - 1}$ ; since $p ∣ ∣ G ∣$ and $p - 1 \geq 1$ , one has $p ∣ ∣ Ω ∣$ .

L3L4L5construct

2.1

Let $1 \in Z / p$ act on $Ω$ by cyclic rotation. If $g_{0} \dots g_{p - 1} = e$ , then $g_{1} \dots g_{p - 1} g_{0} = g_{0}^{- 1} (g_{0} \dots g_{p - 1}) g_{0} = e$ , so rotation preserves $Ω$ ; $p$ rotations are the identity, and [L2] therefore gives an action of the finite $p$ -group $Z / p$ .

step 1.1L2

3.1

A tuple is fixed by every rotation exactly when it is constant, say $(g, \dots, g)$ , and it lies in $Ω$ exactly when $g^{p} = e$ .

step 2.1L6

3.2

By [L1], $∣ Ω ∣ \equiv ∣ Ω^{Z / p} ∣ (m o d p)$ . Step 1.1 makes the left side divisible by $p$ , so the number of fixed tuples is divisible by $p$ . The constant identity tuple is fixed, and a positive multiple of $p > 1$ cannot equal $1$ , so there is another fixed tuple.

step 1.1step 2.1L1L4L5

4.1

By step 3.1, this second tuple is $(g, \dots, g)$ for some $g \neq e$ with $g^{p} = e$ . By [L6], the positive order of $g$ divides the prime $p$ and is not $1$ , so it is $p$ .

step 3.1step 3.2L4L6discharge-construct∎

Depends on

Used by

Dependency tree · two levels

67 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

K. Conrad, Group Actions, Theorem 5.4 (standard reference, not scraped)
T. W. Judson, Abstract Algebra: Theory and Applications, 14.2 (standard reference, not scraped)

Cauchy's theorem: if a prime pp divides ∣G∣∣G∣, then GG has an element of order pp