Groups, Subgroups, and Cosets
A group is the algebraic distillation of a single idea: symmetry. Strip a set of transformations down to the four features every family of symmetries shares — you can compose two of them, the composition does not care how you bracket it, there is a "do nothing" move, and every move can be undone — and you are left with the group axioms. Nothing about squares, molecules, or particles survives the abstraction, yet the abstraction is exactly what makes group theory so powerful: one theorem proved about the bare axioms simultaneously constrains the vibrational modes of ammonia, the selection rules of an atomic transition, and the multiplet structure of the quark model. This article develops that abstract skeleton and then dissects it. The dissection tool is the subgroup — a self-contained smaller group nested inside the larger one — together with the cosets it generates: the shifted copies of the subgroup that tile the parent set into equal, non-overlapping blocks. The single most important consequence, and the first non-trivial theorem in all of group theory, is Lagrange's: because every block has the same size, the order of the whole factorizes as index times subgroup order, and so the order of any subgroup must divide the order of the group. From that one arithmetic fact a cascade follows — the order of every element divides the group order, groups of prime order can only be cyclic, and Euler's centuries-old number-theoretic congruence drops out as a special case. The coset construction is also the doorway to everything downstream: normal subgroups, quotient groups, homomorphisms, and the representation theory on which the physics of symmetry is built.
- A group (G, ·) is a set with an associative product that has an identity and in which every element is invertible — the minimal structure that captures "a family of symmetries you can compose and undo".
- A subgroup H ≤ G is a non-empty subset closed under the product and inverses; the one-step test ab⁻¹ ∈ H checks both at once.
- Left cosets gH partition G into disjoint blocks all of the same size |H|; the number of blocks is the index [G : H].
- Lagrange's theorem is the immediate payoff: |G| = [G : H] · |H|, so every subgroup order divides |G|.
- Consequences cascade: ord(g) divides |G|, groups of prime order are forced to be cyclic, and Euler's congruence a^φ(n) ≡ 1 is just g^|G| = e in disguise.
- The converse is false — a divisor of |G| need not occur as a subgroup order (A₄ has order 12 but no subgroup of order 6).
- Key idea: a subgroup H slices its parent group G into equal-sized, mutually disjoint cosets; counting the slices gives Lagrange's theorem |G| = [G : H] · |H|, which forces subgroup orders — and hence element orders — to divide |G|.
Where this lives today
The coset arithmetic that Lagrange launched is quietly load-bearing across modern physics and computation. In molecular and solid-state physics, the finite symmetry group of a molecule or crystal — its point group — has an order equal to the number of symmetry operations, and Lagrange's theorem restricts which subgroups can appear as the residual symmetry after a distortion or a symmetry-lowering phase transition; the coset decomposition of the point group with respect to a site-symmetry subgroup is exactly what enumerates the equivalent atomic positions in a Wyckoff analysis. In quantum mechanics the symmetry group of a Hamiltonian dictates selection rules, and the fact that ord(g) divides |G| pins the eigenvalues of a symmetry operator to roots of unity — a rotation by 2π/n has eigenvalues that are nth roots of unity, giving the discrete quantum numbers of the states. The most visible modern descendant, however, is cryptographic: RSA and Diffie–Hellman rest directly on Euler's corollary a^φ(n) ≡ 1, which makes modular exponentiation in (ℤ/nℤ)^× invertible; the security of the entire public-key edifice is a statement about the order of that finite group and the divisibility relations Lagrange supplies. And in the elementary-particle world, the classification of small groups that Lagrange's constraints make possible is what lets a physicist recognize, for example, the C₃ᵥ symmetry of ammonia as abstractly the very S₃ dissected in this article.
The physical picture
Picture a physical object and the complete set of moves that leave it looking exactly the same. For an equilateral triangle those moves are: do nothing, rotate by ±120°, and reflect across any of the three axes through a vertex — six moves in all. Each is reversible (there is an inverse), doing nothing is one of them (the identity), and any two done in succession give a third move on the list (closure). Associativity is the quiet axiom: when you chain three moves, the mental bracketing of the chain is irrelevant, because each move is a function on the corners and function composition is always associative. These six moves are the group S₃, and the first figure lays them out and highlights a two-element self-contained sub-collection — a subgroup.
A subgroup is a sub-collection of moves that is self-sufficient: combine or undo its members and you never escape it. To see how a subgroup organizes the whole group, prepend one outside move to every element of H. This produces a coset gH — a shifted copy of H — and the remarkable fact is that these copies never overlap and always have exactly |H| members, like cards dealt from a deck into equal hands. The group is thereby tiled into equal blocks, and simply counting the tiles is Lagrange's theorem. The second figure makes the tiling literal for S₃: the six elements are laid out as a rectangle whose rows are the three cosets of H = {e, (1 2)}, each row exactly two cells wide.
The core equations
The definition packs four demands into one line. A group is a set G with a binary operation · that is closed and associative, contains a two-sided identity, and gives every element a two-sided inverse. The abbreviations (G1)–(G4) below are the entire axiom system; everything else in group theory is a theorem built from them.
where: — the underlying set; — the operation, usually written by juxtaposition ; — the identity; — the inverse of ; the quantifiers range over all elements of . Reading: closure keeps you inside , associativity lets you drop brackets, the identity does nothing, and inverses let you undo. One does not need to assume or are unique — that follows from the axioms in two lines. Testing whether a subset is itself a group need not re-check all four axioms. Associativity is automatic (it is inherited from ), so it suffices to confirm the subset is non-empty and closed under "divide", i.e. under forming . This single condition secretly delivers the identity, inverses, and closure at once.
where: — a candidate subset; reads " is a subgroup of ". Reading: put to get ; put to get ; then recovers closure — so the one-step test is genuinely equivalent to the full subgroup definition. Given a subgroup, its left coset by is the shifted copy , and two elements land in the same coset exactly when they differ by a member of on the left. That "differ by an element of " rule is an equivalence relation, and its equivalence classes are precisely the cosets — which is why cosets partition the group cleanly.
where: — the left coset with representative ; — the equivalence relation whose classes are the left cosets . Reading: any element of a coset can serve as its representative, and distinct cosets are disjoint — there is no partial overlap, only "identical" or "share nothing". Counting the equal-sized cosets gives the central theorem. With the number of distinct cosets (the index) and their common size, the order of a finite group factorizes exactly.
where: — the order (number of elements) of ; — the order of the subgroup; — the index. Reading: because and are positive integers, must divide — subgroup orders are constrained by the arithmetic of alone. Applying Lagrange to the cyclic subgroup generated by a single element turns the theorem into a statement about element orders, and raising to that order gives a universal identity every element satisfies.
where: — the smallest with , equal to ; — the cyclic subgroup generated by . Reading: because is a subgroup, its size divides , and since is a multiple of , collapses to . Specializing to the multiplicative group of residues coprime to — whose order is Euler's totient — reproduces a landmark of number theory with no arithmetic beyond Lagrange.
where: — the number of integers in coprime to , i.e. ; — any integer coprime to . Reading: this is for the group of units mod ; for prime it becomes Fermat's little theorem .
The goal is to prove that for any finite group G and subgroup H the order factorizes as |G| = [G : H] · |H|, and to do so from the ground up: we build the equivalence relation whose classes are the left cosets, verify it really is an equivalence relation, show every class is a shifted copy of H of exactly the same size, and finally add the block sizes. The whole argument rests on two facts — that cosets are either identical or disjoint, and that left-multiplication by a fixed element is a bijection — and it is the universally cited route in Stone and Goldbart, Arfken–Weber, and Dummit and Foote.
Step 1. Define a relation on G by a ~ b ⟺ a⁻¹b ∈ H. Check the three properties: reflexive because a⁻¹a = e ∈ H; symmetric because if a⁻¹b ∈ H then its inverse (a⁻¹b)⁻¹ = b⁻¹a ∈ H; transitive because (a⁻¹b)(b⁻¹c) = a⁻¹c ∈ H whenever both factors lie in H. So ~ is an equivalence relation. Why: equivalence relations automatically carve a set into disjoint classes that cover it, which is exactly the partition we want — we get "no overlaps" for free rather than proving it by hand.
Step 2. Identify the class of a. By definition [a] = { b : a⁻¹b ∈ H } = { b : b = ah for some h ∈ H } = aH, the left coset. Hence the left cosets are the equivalence classes, so distinct cosets are disjoint and their union is all of G. Why: this converts the abstract classes of Step 1 into the concrete, computable objects gH — the shifted copies you can actually list out, as in the S₃ rectangle.
Step 3. Show every coset has |H| elements. Fix a and define φ : H → aH by φ(h) = ah. It is surjective by the definition of aH, and injective because ah₁ = ah₂ implies h₁ = h₂ after left-multiplying by a⁻¹ (left-cancellation, valid because a is invertible). A bijection forces |aH| = |H| for every a. Why: this is where invertibility does the real work — cancellation is the reason the rows of the rectangle cannot be ragged, so all blocks are genuinely the same size.
Step 4. Add the blocks. G is the disjoint union of [G : H] cosets, each of size |H|, so |G| = Σ |aH| = [G : H] · |H|. In particular |H| divides |G|, and applying this to H = ⟨g⟩ gives ord(g) | |G|. Why: the counting step is trivial only because Steps 1–3 guaranteed equal, non-overlapping blocks; the divisibility of |G| by every subgroup order — the theorem's real force — is now immediate.
Go deeper · Derivation B · 1 minFrom Lagrange to the Euler and Fermat congruences
The aim is to show that a purely arithmetic fact known since the eighteenth century — Euler's congruence a^φ(n) ≡ 1 (mod n) — is nothing more than the corollary g^|G| = e read inside one particular finite group, the group of units modulo n. We first assemble that group and count it, then apply Lagrange's corollary verbatim, then collapse to Fermat's little theorem for prime moduli. This is the cleanest demonstration that group theory subsumes elementary number theory rather than merely resembling it.
Step 1. Assemble the group of units (ℤ/nℤ)^× = { ā : gcd(a, n) = 1 }, the residues coprime to n under multiplication mod n. It is closed (a product of two numbers coprime to n is coprime to n) and every element has an inverse (Bézout's identity gives ax + ny = 1, so x̄ is the inverse of ā). Its order is Euler's totient, |(ℤ/nℤ)^×| = φ(n), by definition of φ. Why: verifying closure and inverses is exactly checking the group axioms hold, so Lagrange's machinery is licensed to act on this set.
Step 2. Apply the corollary g^|G| = e. With G = (ℤ/nℤ)^×, identity 1̄, and |G| = φ(n), the identity reads ā^φ(n) = 1̄, i.e. a^φ(n) ≡ 1 (mod n) for every a coprime to n. Why: no arithmetic input beyond Lagrange is needed — the congruence is the group identity translated back into the language of residues.
Step 3. Specialize to a prime. For n = p prime, every one of 1, …, p−1 is coprime to p, so φ(p) = p − 1, and the congruence becomes a^(p−1) ≡ 1 (mod p) for p ∤ a — Fermat's little theorem. As a numerical check, take p = 5: φ(5) = 4 and 2⁴ = 16 = 3·5 + 1 ≡ 1 (mod 5). Why: recovering a seventeenth-century theorem of Fermat as a one-line special case is the sharpest evidence that the abstract order-divides-order fact carries real arithmetic content.
Go deeper · Derivation C · 2 minCosets and non-normality in the symmetric group S₃
The goal is to compute, entirely by hand, the left and right cosets of a concrete order-2 subgroup inside the six-element group S₃, to confirm Lagrange's factorization 6 = 3 × 2 explicitly, and — crucially — to exhibit a subgroup whose left cosets differ from its right cosets. That last point is the whole reason normal subgroups are singled out later: coset partitions from the left and from the right need not agree, and only when they do can the cosets themselves be multiplied to form a quotient group. Every permutation product below is composed right-to-left, the standard convention.
Step 1. List the group. S₃ has the six elements e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2) — the identity, three transpositions, and two 3-cycles, so |S₃| = 3! = 6. Take H = ⟨(1 2)⟩ = {e, (1 2)}, of order 2. Why: a concrete, fully enumerated group is the only way to see cosets as actual sets of elements rather than as a formula.
Step 2. Build the left cosets gH. The identity gives H itself. Compute (1 3)(1 2): apply (1 2) then (1 3), giving 1 → 2, 2 → 3, 3 → 1 = (1 2 3), so (1 3)H = {(1 3), (1 2 3)}. Compute (2 3)(1 2) = (1 3 2), so (2 3)H = {(2 3), (1 3 2)}. The three left cosets are disjoint, each of size 2, and cover S₃. Why: this is the S₃ rectangle of the second figure realized by explicit multiplication — the equal, non-overlapping rows are computed, not assumed.
Step 3. Build the right coset H(1 3) and compare. Compute (1 2)(1 3): apply (1 3) then (1 2), giving 1 → 3, 2 → 1, 3 → 2 = (1 3 2), so H(1 3) = {(1 3), (1 3 2)}. But the left coset (1 3)H = {(1 3), (1 2 3)}. Since (1 3 2) ≠ (1 2 3), the right coset differs from the left coset: H is not normal. Why: this single inequality is the physical content of "non-normal" — the left tiling and the right tiling of the same group genuinely disagree, so no consistent quotient product can be defined on these cosets.
Step 4. Confirm Lagrange and contrast with A₃. Index times order gives 3 × 2 = 6 = |S₃|, exactly. By contrast the order-3 subgroup A₃ = ⟨(1 2 3)⟩ has index 6/3 = 2, and any index-2 subgroup is automatically normal because its two cosets must be H and G ∖ H on both sides. Why: seeing one non-normal and one normal subgroup in the same small group shows that normality is an extra condition, not a consequence of being a subgroup.
Limiting cases
The coset and Lagrange machinery specializes sharply according to the group at hand; the table collects the boundary and special cases that recur constantly in physics.
| Case | Result | Meaning | ||||
|---|---|---|---|---|---|---|
| Abelian group | gH = Hg for all g | Every subgroup is normal, left and right cosets coincide, and every quotient G/H exists. Example: ℤₙ and its subgroups dℤₙ for each divisor d of n. | ||||
| Prime order \ | G\ | = p | H ∈ {{e}, G} | Lagrange forbids any proper non-trivial subgroup, forcing G ≅ ℤ_p — cyclic and abelian. The simplest classification theorem in group theory. | ||
| Index two, [G : H] = 2 | gH = G ∖ H = Hg | A subgroup of index 2 is automatically normal: both the left and right partitions can only be {H, G ∖ H}. Example: Aₙ is normal in Sₙ. | ||||
| Trivial vs. whole subgroup | [G : {e}] = \ | G\ | , [G : G] = 1 | The two extreme subgroups give index \ | G\ | (singleton cosets) and index 1 (a single coset) — the endpoints of every coset partition. |
Worked examples
1 · Cosets, index, and element orders in ℤ₁₂ (fully worked)
Take the cyclic group G = ℤ₁₂ = {0, 1, …, 11} under addition mod 12, and the subset H = {0, 4, 8}. Show H is a subgroup, find its cosets and index, confirm Lagrange, and check that every element order divides 12.
Step 1. Confirm H is a subgroup with the one-step test (written additively as a − b ∈ H). The differences among {0, 4, 8} are 0, ±4, ±8, and mod 12 these are 0, 4, 8 — all back in H. So H ≤ ℤ₁₂ with |H| = 3.
Step 2. Form the cosets g + H. Starting from 0, 1, 2, 3 gives 0 + H = {0, 4, 8}, 1 + H = {1, 5, 9}, 2 + H = {2, 6, 10}, 3 + H = {3, 7, 11}. Every element of ℤ₁₂ appears once, so these four disjoint cosets exhaust the group and the index is [ℤ₁₂ : H] = 4.
Step 3. Confirm Lagrange: |G| = [G : H] · |H| = 4 × 3 = 12, exactly |ℤ₁₂|.
Step 4. Check element orders against ord(g) | |G|. In ℤₙ the order of g is n/gcd(n, g): ord(3) = 12/gcd(12, 3) = 12/3 = 4, ord(8) = 12/gcd(12, 8) = 12/4 = 3, ord(5) = 12/gcd(12, 5) = 12/1 = 12. Every one of 4, 3, 12 divides 12, as Lagrange demands, and the generator 5 (order 12) confirms ℤ₁₂ is cyclic. Cosets: 4 of them; Lagrange 4 × 3 = 12; all element orders divide 12.
2 · Euler's theorem and the order of an element in (ℤ/9ℤ)^×
The units mod 9 are (ℤ/9ℤ)^× = {1, 2, 4, 5, 7, 8}. Verify Euler's congruence for a = 2, find the order of 2, and use Lagrange to explain what orders are even possible.
Step 1. Count the group: the residues coprime to 9 are 1, 2, 4, 5, 7, 8, so φ(9) = 6 and |(ℤ/9ℤ)^×| = 6.
Step 2. Verify a^φ(n) ≡ 1. Compute powers of 2 mod 9: 2, 4, 8, then 2⁴ = 16 ≡ 7, 2⁵ = 32 ≡ 5, 2⁶ = 64 = 7·9 + 1 ≡ 1 (mod 9). So 2⁶ ≡ 1, exactly Euler's theorem with φ(9) = 6.
Step 3. Read off the order. The powers 2, 4, 8, 7, 5, 1 are all six distinct units and 6 is the first exponent returning 1, so ord(2) = 6. Since ord(2) = |G|, the element 2 generates the whole group — it is a primitive root mod 9, and (ℤ/9ℤ)^× ≅ ℤ₆ is cyclic.
Step 4. Use Lagrange to bound the possibilities. Any element order must divide φ(9) = 6, so the only possible orders are 1, 2, 3, 6 — and indeed 1 has order 1, 8 has order 2 (8² = 64 ≡ 1), 4 and 7 have order 3, and 2, 5 have order 6. 2⁶ ≡ 1 mod 9, ord(2) = 6, and every element order lies in {1, 2, 3, 6}, each dividing 6.
Physics of the portrait
Lagrange's theorem carries the name of a man who never saw a group. Working half a century before the abstract concept existed, Joseph-Louis Lagrange counted how many values a function of a polynomial's roots can take when the roots are permuted — and stumbled onto the divisibility fact that now anchors the whole subject.
Origins · 1 minJoseph-Louis Lagrange and how this was discovered
Joseph-Louis Lagrange — born Giuseppe Luigi Lagrangia in Turin in 1736 — was one of the towering analysts of the eighteenth century, remembered by physicists chiefly for Lagrangian mechanics and the calculus of variations. His connection to group theory is older and subtler. In his 1770–71 memoir Réflexions sur la résolution algébrique des équations, Lagrange asked why the general quintic resisted the solution-by-radicals that worked for degrees two, three, and four. His method was to track a rational function of the roots and count how many distinct values it takes as the roots are permuted, and he found that this number always divides n!, the total number of permutations. That observation — that the size of an orbit of permutations divides the size of the full permutation group — is the concrete seed of what we now call Lagrange's theorem, even though Lagrange had neither the word "group" nor the axioms. The abstraction came later and from others: Paolo Ruffini and Niels Abel used permutation arguments to prove the quintic unsolvable; Évariste Galois, around 1830, introduced the term groupe and the pivotal notion of a normal subgroup; Augustin-Louis Cauchy systematized permutation groups; and Arthur Cayley in 1854 finally gave the abstract axiomatic definition, freeing "group" from permutations entirely. The theorem that bears Lagrange's name is thus a nineteenth-century generalization of an eighteenth-century counting insight — a fitting emblem of how the deepest structural facts often appear first as arithmetic coincidences.
Going further
Texts. Stone and Goldbart, Mathematics for Physics, chapter on group theory · Arfken, Weber and Harris, Mathematical Methods for Physicists, 7th ed., ch. 17 · Riley, Hobson and Bence, Mathematical Methods for Physics and Engineering, 3rd ed., ch. 28 · Hamermesh, Group Theory and Its Application to Physical Problems · Tinkham, Group Theory and Quantum Mechanics · Dummit and Foote, Abstract Algebra, 3rd ed., ch. 1–3. Key anchor results: the ℤ₁₂ cosets are {0,4,8}, {1,5,9}, {2,6,10}, {3,7,11} with index 4; the powers of 2 mod 9 give order 6 and 2⁶ ≡ 1; φ(5) = 4 with 2⁴ ≡ 1 mod 5; and the A₄ counterexample has order 12 with no subgroup of order 6. Standard values: |S₃| = 6, φ(p) = p − 1, |A₄| = 12.
Prerequisites. Sets, functions, and equivalence relations · Elementary number theory (divisibility, primes) · Permutations and combinatorics. Leads to. Conjugacy Classes and Normal Subgroups · Group Homomorphisms and Quotient Groups · Group Representations · Lie Groups · The Rotation Group SO(3).
Ask the tutor: Ask me to walk through the coset partition of any small group — Sₙ, ℤₙ, a dihedral or point group — and watch Lagrange's 6 = 3 × 2 fall out element by element.
What is the difference between a subgroup and a coset?
A subgroup H is itself a group: it contains the identity, is closed under the operation, and holds the inverse of each of its elements. A coset gH (for g not in H) is generally not a group — it does not even contain the identity unless g already lies in H. The coset is just the set you get by multiplying every element of H on the left by the fixed element g, a "shifted copy" of H. The subgroup H is special because it is the one coset that is also a subgroup (namely eH = H); all the other cosets are equal-sized translates of it. Together, the cosets partition the group into disjoint blocks, and that partition — not any individual coset's internal structure — is what Lagrange's theorem counts.
Why does associativity not need to be checked for a subgroup?
Associativity is a property of the operation, and a subgroup uses the very same operation as its parent group. If (ab)c = a(bc) holds for all elements of G, it certainly still holds when a, b, c happen to be drawn from a subset H ⊆ G — nothing about restricting the domain can break an identity that was already universally true. This is why the subgroup tests only ask about closure, the identity, and inverses (or, in the compact one-step form, only that ab⁻¹ ∈ H): those are the properties that can fail for a subset even when they hold for the whole group. Associativity, being inherited automatically, never has to be re-verified.
Does Lagrange's theorem work for infinite groups?
Not in its counting form. The statement |G| = [G : H] · |H| is about finite cardinalities, and for infinite groups these orders are infinite, so the arithmetic "|H| divides |G|" is meaningless. However, the structural heart of the proof survives: the left cosets of any subgroup — finite or not — still partition the group into disjoint blocks of equal cardinality, and the index [G : H] (the number of cosets) is still well defined, possibly infinite. So the coset partition is completely general; only the divisibility corollary is special to finite groups. For infinite groups one instead studies the index directly — for instance, the even integers have index 2 in ℤ, a perfectly meaningful finite index inside an infinite group.
What exactly makes a subgroup "normal", and why single it out?
A subgroup H is normal when its left and right cosets coincide, gH = Hg for every g. Equivalently, conjugation gHg⁻¹ maps H back to itself. The reason to single this out is that only for a normal subgroup can you multiply cosets consistently: the rule (aH)(bH) = (ab)H is well defined — independent of which representatives a and b you pick — precisely when H is normal. When it holds, the set of cosets G/H becomes a group in its own right, the quotient group, which is the gateway to homomorphisms and the isomorphism theorems. The S₃ example in this article shows a non-normal subgroup, where the left and right tilings disagree and no such quotient product exists.
If a divisor of |G| need not give a subgroup, what can I actually rely on?
Rely on the one-directional statement: every subgroup order divides |G|, so any candidate order that fails to divide |G| is immediately impossible (a group of order 15 can have no subgroup of order 4, since 4 does not divide 15). For the reverse direction, use the correct partial converses rather than the false full converse. Cauchy's theorem guarantees that for every prime p dividing |G| there is a subgroup — indeed an element — of order p. The Sylow theorems go further, guaranteeing subgroups of order equal to the largest power of each prime dividing |G|, and controlling how many there are. It is only the intermediate composite divisors (like 6 in A₄ of order 12) that can fail to appear.
How is the abstract group S₃ the same as the triangle's symmetry group C₃ᵥ?
They are isomorphic — there is a one-to-one correspondence between their elements that respects composition. The six rigid symmetries of an equilateral triangle (identity, two rotations, three mirror reflections) act by permuting the three labelled corners, and every permutation of three objects arises exactly once this way. So the map "symmetry ↦ its permutation of the corners" is a bijection from C₃ᵥ onto S₃ that turns composition of symmetries into composition of permutations. Because all the group-theoretic content — orders, subgroups, cosets, normality — is preserved by an isomorphism, anything proved about S₃ (its order-2 subgroups are non-normal, its order-3 subgroup A₃ is normal) is automatically a fact about the triangle's symmetries, and about the ammonia molecule that shares the same symmetry group.