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Lagrange's Theorem: Why Subgroup Orders Divide Group Orders

We prove Lagrange's theorem — the fundamental result that the order of a subgroup divides the order of the group — and explore its many consequences, from Fermat's little theorem to the structure theory of finite groups.

The Theorem

Lagrange's Theorem

If GG is a finite group and HGH \leq G is a subgroup, then H|H| divides G|G|. Moreover:

G=H[G:H]|G| = |H| \cdot [G : H]

where [G:H][G : H] is the index of HH in GG — the number of left cosets of HH.

This is one of the most fundamental theorems in algebra. It constrains which subgroups a group can have and is the starting point for much of finite group theory.


Cosets

The proof rests on the concept of cosets.

Definition

For gGg \in G, the left coset of HH containing gg is:

gH={gh:hH}gH = \{gh : h \in H\}

Similarly, the right coset is Hg={hg:hH}Hg = \{hg : h \in H\}.

Key Properties

Lemma. Let HGH \leq G be a subgroup of a finite group.

  1. Every coset gHgH has exactly H|H| elements.
  2. Two cosets aHaH and bHbH are either equal or disjoint.
  3. GG is the disjoint union of its left cosets.

Proof of (1). The map φ:HgH\varphi: H \to gH defined by φ(h)=gh\varphi(h) = gh is a bijection: it is clearly surjective, and if gh1=gh2gh_1 = gh_2 then h1=h2h_1 = h_2 (multiply by g1g^{-1}). So gH=H|gH| = |H|. \square

Proof of (2). Suppose aHbHaH \cap bH \neq \emptyset. Then there exist h1,h2Hh_1, h_2 \in H with ah1=bh2ah_1 = bh_2, giving a=bh2h11bHa = bh_2h_1^{-1} \in bH. For any ahaHah \in aH, we have ah=bh2h11hbHah = bh_2h_1^{-1}h \in bH, so aHbHaH \subseteq bH. By symmetry, bHaHbH \subseteq aH. \square

Proof of (3). Every gGg \in G belongs to gHgH (since eHe \in H, so g=gegHg = ge \in gH). By (2), distinct cosets are disjoint. \square


Proof of Lagrange's Theorem

Proof.

By the lemma above, the left cosets of HH partition GG into disjoint subsets, each of size H|H|:

G=g1Hg2HgkHG = g_1 H \sqcup g_2 H \sqcup \cdots \sqcup g_k H

where k=[G:H]k = [G:H] is the number of distinct cosets. Counting elements:

G=kH=[G:H]H|G| = k \cdot |H| = [G:H] \cdot |H|

Therefore H|H| divides G|G|. \square

The proof is remarkably simple, yet its consequences are deep.


Example

Consider G=S3G = S_3 (the symmetric group on 3 elements), which has order S3=6|S_3| = 6.

The subgroup H={e,(12)}H = \{e, (12)\} has order 22. By Lagrange's theorem, [S3:H]=6/2=3[S_3 : H] = 6/2 = 3.

The three left cosets are:

eH={e,(12)},(13)H={(13),(132)},(23)H={(23),(123)}eH = \{e, (12)\}, \quad (13)H = \{(13), (132)\}, \quad (23)H = \{(23), (123)\}

These partition S3S_3 into three disjoint pairs.


Immediate Consequences

Corollary 1: Order of Elements Divides Group Order

If GG is a finite group and gGg \in G, then ord(g)\operatorname{ord}(g) divides G|G|.

Proof. The cyclic subgroup g={e,g,g2,,gk1}\langle g \rangle = \{e, g, g^2, \ldots, g^{k-1}\} has order k=ord(g)k = \operatorname{ord}(g). By Lagrange's theorem, kGk \mid |G|. \square

Corollary 2: gG=eg^{|G|} = e

For every gg in a finite group GG:

gG=eg^{|G|} = e

since G=km|G| = k \cdot m and gkm=(gk)m=em=eg^{km} = (g^k)^m = e^m = e.

Corollary 3: Groups of Prime Order Are Cyclic

If G=p|G| = p is prime, then GG has no proper nontrivial subgroups (since the only divisors of pp are 11 and pp). Therefore G=gG = \langle g \rangle for any geg \neq e, and GZ/pZG \cong \mathbb{Z}/p\mathbb{Z}.


Fermat's Little Theorem

Fermat's Little Theorem. If pp is prime and gcd(a,p)=1\gcd(a, p) = 1, then:

ap11(modp)a^{p-1} \equiv 1 \pmod{p}

Proof via Lagrange. The group (Z/pZ)={1,2,,p1}(\mathbb{Z}/p\mathbb{Z})^* = \{1, 2, \ldots, p-1\} under multiplication modulo pp has order p1p - 1. By Corollary 2:

ap11(modp)a^{p-1} \equiv 1 \pmod{p} \qquad \square

Euler's Generalization

More generally, (Z/nZ)(\mathbb{Z}/n\mathbb{Z})^* has order φ(n)\varphi(n) (Euler's totient), so:

aφ(n)1(modn)whenever gcd(a,n)=1a^{\varphi(n)} \equiv 1 \pmod{n} \quad \text{whenever } \gcd(a, n) = 1

This is Euler's theorem, also an immediate consequence of Lagrange's theorem.


The Converse Is False

Lagrange's theorem does not have a converse. If dd divides G|G|, there need not exist a subgroup of order dd.

Example: A4A_4 has order 1212, and 6126 \mid 12, but A4A_4 has no subgroup of order 66.

However, partial converses exist:

  • Cauchy's Theorem: If a prime pp divides G|G|, then GG has an element of order pp (hence a subgroup of order pp).
  • Sylow's Theorems: If pkp^k divides G|G| (with pp prime), then GG has a subgroup of order pkp^k.

Cauchy's Theorem

Cauchy's Theorem. If pp is prime and pGp \mid |G|, then GG contains an element of order pp.

Proof (McKay). Let S={(g1,,gp)Gp:g1g2gp=e}S = \{(g_1, \ldots, g_p) \in G^p : g_1 g_2 \cdots g_p = e\}. Then S=Gp1|S| = |G|^{p-1} (choose g1,,gp1g_1, \ldots, g_{p-1} freely; gpg_p is determined). The cyclic group Z/pZ\mathbb{Z}/p\mathbb{Z} acts on SS by cyclic permutation. Each orbit has size 11 or pp. The fixed points are tuples (g,g,,g)(g, g, \ldots, g) with gp=eg^p = e. Since S=Gp10(modp)|S| = |G|^{p-1} \equiv 0 \pmod{p} and the identity contributes one fixed point, the number of fixed points is 0(modp)\equiv 0 \pmod{p}, so there are at least p1p - 1 more fixed points, giving geg \neq e with gp=eg^p = e. \square


The Index Formula

Tower Law. If KHGK \leq H \leq G are finite groups, then:

[G:K]=[G:H][H:K][G : K] = [G : H] \cdot [H : K]

Proof. G=[G:H]H=[G:H][H:K]K|G| = [G:H] \cdot |H| = [G:H] \cdot [H:K] \cdot |K|, and G=[G:K]K|G| = [G:K] \cdot |K|. \square

This is the group-theoretic analogue of the tower law for field extensions: [L:K]=[L:F][F:K][L:K] = [L:F][F:K].


Applications

RSA Cryptography

The RSA algorithm relies on Euler's theorem: for n=pqn = pq and gcd(a,n)=1\gcd(a, n) = 1:

aφ(n)1(modn)a^{\varphi(n)} \equiv 1 \pmod{n}

Choosing e,de, d with ed1(modφ(n))ed \equiv 1 \pmod{\varphi(n)} gives (ae)d=aed=a1+kφ(n)=a(aφ(n))ka(a^e)^d = a^{ed} = a^{1 + k\varphi(n)} = a \cdot (a^{\varphi(n)})^k \equiv a, enabling encryption and decryption.

Counting Group Homomorphisms

Lagrange's theorem constrains homomorphisms: if ϕ:GH\phi: G \to H is a homomorphism, then Im(ϕ)|\operatorname{Im}(\phi)| divides both G|G| and H|H|. If gcd(G,H)=1\gcd(|G|, |H|) = 1, the only homomorphism is trivial.

Burnside's Lemma

In counting orbits under group actions, Lagrange's theorem ensures the orbit-stabilizer theorem: G=Orb(x)Stab(x)|G| = |\operatorname{Orb}(x)| \cdot |\operatorname{Stab}(x)|, which is a direct application.


Historical Note

Despite its name, Lagrange's theorem in its modern form was not stated by Lagrange himself. Joseph-Louis Lagrange (1770–71) proved the special case that the order of a permutation divides n!n! in his work on polynomial equations. The general theorem emerged from the work of Galois and Cauchy in the early 19th century.


Summary

HG    HGG=[G:H]H(coset counting)Consequences:ord(g)G,gG=eG=p    GZ/pZap11(modp)(Fermat)aφ(n)1(modn)(Euler)\begin{aligned} &H \leq G \implies |H| \mid |G| \\[6pt] &|G| = [G:H] \cdot |H| \quad \text{(coset counting)} \\[8pt] &\text{Consequences:} \\[4pt] &\quad \operatorname{ord}(g) \mid |G|, \quad g^{|G|} = e \\[4pt] &\quad |G| = p \implies G \cong \mathbb{Z}/p\mathbb{Z} \\[4pt] &\quad a^{p-1} \equiv 1 \pmod{p} \quad \text{(Fermat)} \\[4pt] &\quad a^{\varphi(n)} \equiv 1 \pmod{n} \quad \text{(Euler)} \end{aligned}

References