Lagrange's Theorem: Why the Order of a Subgroup Always Divides the Order of the Group

The precise statement

Let G be a finite group and let H ≤ G be a subgroup (a subset closed under the group operation and inverses, containing the identity). Lagrange's Theorem asserts:

• The order |H| divides |G|.
• More precisely, |G| = [G:H] · |H|, where the index [G:H] is the number of distinct left cosets of H in G.

A left coset is a set gH = { gh : h ∈ H } for some g ∈ G. The theorem does not require G to be abelian, nor H to be normal. The only essential hypothesis is finiteness of G — that is what makes |G| a well-defined natural number for divisibility to even mean something. In the infinite setting the index formula |G| = [G:H]·|H| still holds as an equation of cardinals, but 'divides' loses its arithmetic bite.

The picture: equal-size tiles covering the group

Imagine G as a floor and H as one tile sitting at the identity. Now slide that tile around: for each g ∈ G, the coset gH is a translated copy of the tile H. Two facts make this a clean tiling:

• Every point is covered: each g lies in its own coset gH (since e ∈ H, so g = ge ∈ gH).
• Tiles never partially overlap: two cosets are either identical or disjoint — never sharing just some elements.

So the cosets partition G into non-overlapping blocks. And crucially every tile is the same size as H: the map h ↦ gh is a bijection from H to gH. A finite set partitioned into blocks each of size |H| must have total size a multiple of |H|. That is the whole idea — divisibility is forced by a perfect tiling.

The key idea of the proof (the mechanism)

The engine is a single equivalence relation. Declare a ∼ b iff a⁻¹b ∈ H. Reflexivity uses e ∈ H; symmetry uses closure under inverses; transitivity uses closure under multiplication. The equivalence class of a is exactly the coset aH, so the classes partition G.

Now the load-bearing step: each coset has |H| elements. Fix g and consider φ: H → gH, φ(h) = gh. It is surjective by definition of gH, and injective because gh₁ = gh₂ ⇒ h₁ = h₂ by the cancellation law (multiply on the left by g⁻¹). Injectivity is where the group structure is indispensable — a mere subset would not give cancellation. Hence |gH| = |H| for all g.

Summing over the [G:H] disjoint cosets: |G| = ∑ |gH| = [G:H] · |H|. Divisibility follows immediately. No induction, no deep machinery — just partition plus cancellation.

A worked example: S₃ and its subgroups

Take G = S₃, the symmetric group on {1,2,3}, with |G| = 6. Its elements are the identity e, three transpositions (12), (13), (23), and two 3-cycles (123), (132). Lagrange forces every subgroup order to divide 6, i.e. lie in {1, 2, 3, 6}.

• Order 2 subgroups: each transposition generates one, e.g. H = {e, (12)}. Its left cosets are {e,(12)}, {(13),(132)}, {(23),(123)} — three disjoint pairs, index [G:H] = 3, and 3·2 = 6. ✓
• Order 3 subgroup: A₃ = {e, (123), (132)}, the rotations. Two cosets, index 2, and 2·3 = 6. ✓

Notice what is absent: there is no subgroup of order 4 or 5, because 4 ∤ 6 and 5 ∤ 6. Lagrange rules them out before you even look. Conversely, both allowed divisors 2 and 3 are realized here — but as A₄ shows, being an allowed divisor is not a guarantee.

Why finiteness matters, and the failed converse

The finiteness of G is what turns the tiling into a divisibility statement. Drop it and the arithmetic evaporates: in the infinite group (ℤ, +), the subgroup 2ℤ has index 2 but both 2ℤ and ℤ are infinite, so '|2ℤ| divides |ℤ|' is not a meaningful arithmetic claim — only the cardinal index survives.

The more famous subtlety is the converse, which is false. Lagrange says every subgroup order divides |G|; it does not say every divisor is achieved. The canonical counterexample is the alternating group A₄ of order 12: although 6 divides 12, A₄ has no subgroup of order 6. (Such a subgroup would have index 2, hence be normal, and force a 3-cycle to square into it, a contradiction.) The partial converses that do hold are deeper: Cauchy's theorem (a prime p | |G| ⇒ an element of order p) and the Sylow theorems (prime-power divisors are always realized), plus the full converse for finite abelian groups.

Applications and significance

Lagrange's Theorem is the first structural constraint every group must obey, and it seeds a chain of results:

• Order of an element divides |G|: since ⟨g⟩ is a subgroup of order ord(g), we get ord(g) | |G|, hence g^|G| = e for every g.
• Groups of prime order are cyclic: if |G| = p is prime, any non-identity element has order dividing p, so order p, so generates G. Thus there is exactly one group of order p up to isomorphism.
• Number theory for free: applying g^|G| = e to the multiplicative group (ℤ/nℤ)^× of order φ(n) gives Euler's theorem a^φ(n) ≡ 1 (mod n), and the special case n = p gives Fermat's little theorem a^p−1 ≡ 1 (mod p) — the arithmetic backbone of RSA.

It also underpins the counting behind index calculations, coset enumeration, and the entire theory of quotient groups G/N, whose order is exactly [G:N].

Lagrange's Theorem, its converse, and closely related partial converses

Statement	Direction	True in general?	Where it holds / fails
|H| divides |G| for every subgroup H ≤ G	Lagrange (forward)	Always true	Every finite group
If d divides |G|, then G has a subgroup of order d	Full converse	FALSE	Fails: A₄ (order 12) has no subgroup of order 6
If pᵏ divides |G| (p prime), G has a subgroup of order pᵏ	Sylow / Cauchy	TRUE	Sylow's first theorem; Cauchy for k=1
If d divides |G| and G is a finite abelian group	Converse for abelian	TRUE	Every finite abelian group has a subgroup of each divisor order
ord(g) divides |G| for every element g	Corollary	Always true	Follows: ⟨g⟩ is a subgroup of order ord(g)