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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 $p\mid|G|$ , 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 $|X|\equiv|X^P|\pmod p$ ).

[L2]

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

[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 $\lvert A^{B}\rvert = \lvert A\rvert^{\lvert B\rvert}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $\mathbb{R}$ ).

[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 $a\equiv b\pmod p$ means that $p$ divides $a-b$ , and divisibility means existence of an integer factor (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ , Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ 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 $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · constructive

1.1

Let $\Omega$ be the set of $p$ -tuples $(g_0,\ldots,g_{p-1})\in G^p$ whose ordered product is $e$ . The first $p-1$ coordinates determine the last uniquely as $(g_0\cdots g_{p-2})^{-1}$ , so [L3] gives $|\Omega|=|G|^{p-1}$ ; since $p\mid|G|$ and $p-1\ge1$ , one has $p\mid|\Omega|$ .

L3L4L5construct

2.1

Let $1\in\mathbb Z/p$ act on $\Omega$ by cyclic rotation. If $g_0\cdots g_{p-1}=e$ , then $g_1\cdots g_{p-1}g_0=g_0^{-1}(g_0\cdots g_{p-1})g_0=e$ , so rotation preserves $\Omega$ ; $p$ rotations are the identity, and [L2] therefore gives an action of the finite $p$ -group $\mathbb Z/p$ .

step 1.1L2

3.1

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

step 2.1L6

3.2

By [L1], $|\Omega|\equiv|\Omega^{\mathbb Z/p}|\pmod p$ . 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 $(g,\ldots,g)$ for some $g\ne e$ 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

The subgroup ⟨(1 2 3),(1 2)(3 4)⟩≤ S₄ has order 12 but no subgroup of order 6, so Cauchy's theorem does not extend to composite divisors Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 126 results over 29 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 ppp divides ∣G∣|G|∣G∣, then GGG has an element of order ppp