Cosets and Lagrange's Theorem

How many ways can a finite group be sliced into equal-sized pieces by a subgroup? This question has a startlingly clean answer. It is one of the first truly structural facts a student meets in group theory. Given a subgroup $H$ of a finite group $G$ , the group splits into disjoint blocks. These blocks are the cosets of $H$ , each containing exactly $|H|$ elements. Counting these blocks immediately forces $|H|$ to divide $|G|$ : this is Lagrange's Theorem. The importance of this fact is hard to overstate. It converts a qualitative question ("is $H$ a subgroup of $G$ ?") into a quantitative constraint (" $|H|$ must divide $|G|$ "). This instantly rules out most candidate subgroup orders without any further work. It explains why a group of order 15 cannot contain an element of order 4. It also explains why groups of prime order have no proper nontrivial subgroups at all. It is the seed from which the order-of-an-element theorem grows. Fermat's Little Theorem and eventually the Sylow theorems also grow from it. The big picture is a partition: cosets of $H$ tile $G$ into non-overlapping translates of $H$ . All these translates are the same size. They are like identical tiles covering a floor with no gaps and no overlaps. This geometric picture of an equivalence relation is the true engine behind the theorem. Lagrange's Theorem is really a counting argument dressed up in algebraic language. Once the partition is understood, the divisibility is immediate. This article builds the coset machinery from scratch. It covers the equivalence relation behind cosets, and the proof that every coset has the same cardinality as $H$ . It covers the resulting partition of $G$ , the definition of the index $[G:H]$ , and the full proof of Lagrange's Theorem. It closes with worked examples, and a counterexample showing why finiteness (or an appropriate hypothesis) is needed. Finally, it points toward where cosets reappear — most immediately in the theory of Normal Subgroups and Group Actions.

Key takeaways

For a subgroup $H \le G$ and $g \in G$ , the left coset $gH = \{gh : h \in H\}$ and right coset $Hg = \{hg : h \in H\}$ are translates of $H$ .
Distinct left cosets are either identical or disjoint, so the left cosets partition $G$ .
Every coset of $H$ has exactly the same cardinality as $H$ (via a bijection $h \mapsto gh$ ).
The index $[G:H]$ is the number of distinct left cosets of $H$ in $G$ .
Lagrange's Theorem: if $G$ is finite, $|G| = [G:H]\cdot |H|$ , so $|H|$ divides $|G|$ .
Consequences: element orders divide $|G|$ ; groups of prime order are cyclic with no proper nontrivial subgroups.
Key idea: cosets are equal-sized tiles that partition a group, and counting tiles proves that subgroup order divides group order.

Where this lives today

Cosets and Lagrange's Theorem are not a historical curiosity confined to pure algebra — they recur constantly wherever discrete symmetry or modular structure appears.

Number theory: Fermat's Little Theorem and Euler's Theorem are direct corollaries of Lagrange's Theorem applied to the multiplicative groups $(\mathbb{Z}/p\mathbb{Z})^\times$ and $(\mathbb{Z}/n\mathbb{Z})^\times$ ; this underlies primality testing and the correctness of RSA-style cryptographic schemes.
Cryptography: the security and correctness of Diffie–Hellman key exchange and RSA rely on group orders and the fact that element orders divide the order of the ambient multiplicative group, a direct application of Lagrange's corollary.
Coding theory: cosets of a linear code (a subgroup of $\mathbb{F}_q^n$ ) partition the ambient space into equal-size classes used in syndrome decoding, a direct combinatorial echo of coset partitioning.
Crystallography and physics: the classification of crystallographic symmetry groups uses subgroup indices to organize space groups into point-group cosets, and coset spaces $G/H$ appear as configuration spaces (e.g. spontaneously broken symmetry vacua in particle physics are parametrized by a coset space $G/H$ ).
Computer science / algorithms: coset enumeration (the Todd–Coxeter algorithm) computes the index of a subgroup given by generators and relations, used in computational group theory software.
Combinatorics: counting arguments in enumerative combinatorics (e.g. Burnside's lemma / the orbit-counting theorem in Group Actions) build directly on the coset-counting technique introduced here.

:::quiz[In $S_3$ with $H = \langle(1\,2)\rangle$ , what is $[G:H]$ ?]

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:::faq[What exactly is the index $[G:H]$ when $G$ is infinite?] The index $[G:H]$ is defined as the number of distinct left cosets of $H$ in $G$ , and this definition makes sense even when $G$ is infinite; it can be a finite number or an infinite cardinal. For example, in $G = (\mathbb{Z},+)$ with $H = 2\mathbb{Z}$ , there are only 2 cosets (evens and odds), so $[G:H] = 2$ even though $G$ itself is infinite. In contrast, taking $H = \{0\}$ in the same group gives $[G:H] = |G| = \aleph_0$ . The equal-size-partition part of the coset argument works for infinite groups too; only the clean finite arithmetic equation $|G| = [G:H]\cdot|H|$ (and the resulting integer divisibility conclusion) is specific to the finite case.

Ask the tutor: Can you show me why Cauchy's Theorem gives a partial converse to Lagrange's Theorem for prime divisors?

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Intuition

Think of $G$ as a set of points and $H$ as a "shape" sitting inside it containing the identity $e$ . Left-multiplying every point of $H$ by a fixed $g$ rigidly translates that shape to a new location $gH$ elsewhere in $G$ — the same size and structure, just shifted. Because group multiplication by $g$ is invertible (multiply by $g^{-1}$ to undo it), this translation cannot stretch, shrink, or tear the shape: the translated copy $gH$ has exactly as many elements as $H$ itself. The remarkable fact is that as $g$ ranges over all of $G$ , these translated copies of $H$ never partially overlap: two cosets $gH$ and $g'H$ are either the exact same set or completely disjoint. So $G$ ends up perfectly tiled by non-overlapping, identically-sized copies of $H$ . If $G$ is finite, we can simply count tiles: the number of tiles times the size of each tile equals the size of $G$ . That single sentence is Lagrange's Theorem.

Figure

G

tiled into cosets of

H

: each tile has the same size

|H|

G

\sim_L

|G|

H = \langle g\rangle

|G| = p

H \le G

Worked examples

Example 1: $G = S_3$ , $H = \langle (1\,2) \rangle$ . Let $G = S_3 = \{e, (1\,2), (1\,3), (2\,3), (1\,2\,3), (1\,3\,2)\}$ , the symmetric group on $\{1,2,3\}$ , with $|G| = 6$ . Let $H = \langle (1\,2)\rangle = \{e, (1\,2)\}$ , so $|H| = 2$ . We compute the left cosets. First, $eH = H = \{e, (1\,2)\}$ . Take $g = (1\,3)$ : then

gH = \{(1\,3)\cdot e,\ (1\,3)\cdot(1\,2)\} = \{(1\,3),\ (1\,3)(1\,2)\}.

Composing permutations right-to-left, $(1\,3)(1\,2)$ sends $1\mapsto 2$ (since $(1\,2)$ sends $1\to2$ , then $(1\,3)$ fixes $2$ ), sends $2 \mapsto 1 \mapsto 3$ , and sends $3\mapsto 3\mapsto1$ ; so $(1\,3)(1\,2) = (1\,2\,3)$ . Thus

(1\,3)H = \{(1\,3),\ (1\,2\,3)\}.

Take $g = (2\,3)$ : then $(2\,3)\cdot e = (2\,3)$ and $(2\,3)\cdot(1\,2)$ sends $1\mapsto2\mapsto3$ , $2\mapsto1\mapsto1$ , $3\mapsto3\mapsto2$ , giving $(1\,3\,2)$ . So

(2\,3)H = \{(2\,3),\ (1\,3\,2)\}.

These three sets, $H$ , $(1\,3)H$ , $(2\,3)H$ , are pairwise disjoint and their union is all of $S_3$ (each has 2 elements, $3 \times 2 = 6$ elements total, matching $|G|$ ). So $[G:H] = 3$ , and indeed

|G| = [G:H]\cdot|H| = 3 \cdot 2 = 6.

Note that $(1\,2)H = H$ (since $(1\,2)\in H$ gives the same coset as $e$ ) and $(1\,2\,3)H = (1\,3)H$ (check: $(1\,2\,3)\cdot e = (1\,2\,3)$ and $(1\,2\,3)(1\,2)$ sends $1\to2\to3,\ 2\to1\to2$ ... let's verify directly: apply $(1\,2)$ first then $(1\,2\,3)$ : $1\mapsto2\mapsto3$ , $2\mapsto1\mapsto2$ ... recompute carefully: $(1\,2)$ sends $1\to2,2\to1,3\to3$ ; then $(1\,2\,3)$ sends $1\to2,2\to3,3\to1$ . Composite: $1\to2\to3$ , $2\to1\to2$ , $3\to3\to1$ , giving the permutation $1\to3,2\to2,3\to1$ , i.e. $(1\,3)$ . So $(1\,2\,3)H = \{(1\,2\,3),(1\,3)\} = (1\,3)H$ , confirming consistency.) This illustrates: right cosets need not equal left cosets for this $H$ — for instance $H(1\,3) = \{(1\,3), (1\,2)(1\,3)\}$ , and $(1\,2)(1\,3)$ sends $1\to3\to3$ ... apply $(1,3)$ first: $1\to3,3\to1,2\to2$ ; then $(1,2)$ : $1\to2,2\to1,3\to3$ ; composite $1\to3\to3$ , $2\to2\to1$ , $3\to1\to2$ , giving $(2\,3\,1)$ read as cycle $1\to3,3\to2,2\to1$ , i.e. $(1\,3\,2)$ . So $H(1\,3) = \{(1\,3),(1\,3\,2)\}$ , which is not equal to $(1\,3)H = \{(1\,3),(1\,2\,3)\}$ computed above (they share only $(1\,3)$ ). This shows left and right cosets of a non-normal subgroup genuinely differ — the seed of the distinction that becomes crucial in Normal Subgroups.

Example 2: $G = \mathbb{Z}/12\mathbb{Z}$ , $H = \{0,4,8\}$ . Let $G = \mathbb{Z}/12\mathbb{Z} = \{0,1,\dots,11\}$ under addition mod 12, and let $H = \{0,4,8\}$ (this is a subgroup: it is $\langle 4\rangle$ , closed under addition mod 12 and containing $0$ , with inverses $-4\equiv8$ , $-8\equiv4$ ). Here $|G|=12$ , $|H|=3$ . Since $G$ is abelian, left and right cosets coincide; we just call them cosets. Compute cosets by adding representatives:

0+H = \{0,4,8\}, \qquad 1+H=\{1,5,9\}, \qquad 2+H=\{2,6,10\}, \qquad 3+H=\{3,7,11\}.

Check exhaustiveness: $0+H, 1+H, 2+H, 3+H$ together contain $\{0,4,8\}\cup\{1,5,9\}\cup\{2,6,10\}\cup\{3,7,11\}$ , which is exactly $\{0,1,\dots,11\} = G$ , with no repeats. So there are exactly $[G:H] = 4$ distinct cosets, and

|G| = [G:H]\cdot|H| = 4\cdot 3 = 12,

consistent with Lagrange's Theorem. Notice also that $5 + H = \{5, 9, 1\} = 1+H$ (since $5 - 1 = 4 \in H$ ), illustrating that many different representatives $g$ give the same coset $gH$ exactly when they differ by an element of $H$ — precisely the content of the equivalence relation $\sim_L$ from the proof.

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Scope, edge cases & common traps

Situation	What holds	What can fail
$G$ finite, $H \le G$	$\lvert H\rvert$ divides $\lvert G\rvert$ ; $[G:H] = \lvert G\rvert/\lvert H\rvert$	—
$G$ infinite, $H \le G$	Cosets still partition $G$ into equal-cardinality pieces	" $\lvert H\rvert$ divides $\lvert G\rvert$ " is not meaningful/useful (see counterexample)
$H$ not normal	Left cosets $gH$ and right cosets $Hg$ both partition $G$	$gH \neq Hg$ in general (Example 1)
$n$ divides $\lvert G\rvert$	—	$G$ need not have a subgroup of order $n$ (converse of Lagrange is false in general, e.g. $A_4$ has order 12 but no subgroup of order 6)
$g \in G$	$\lvert g\rvert$ divides $\lvert G\rvert$	$g$ need not have order equal to any particular divisor unless further structure (e.g. cyclicity, Sylow/Cauchy theorems) is known

Common trapThe converse of Lagrange's Theorem is FALSE in general

Lagrange's Theorem says $|H|$ divides $|G|$ whenever $H \le G$ — it does not say that for every divisor $d$ of $|G|$ there exists a subgroup of order $d$ . The alternating group $A_4$ has order $12$ , and $6$ divides $12$ , yet $A_4$ has no subgroup of order 6. (One can verify this by checking that a subgroup of order 6 would have index 2, hence be normal, hence be a union of conjugacy classes including $\{e\}$ ; but the conjugacy class sizes in $A_4$ are $1, 3, 4, 4$ , and no subset of these summing to $6$ containing the class $\{1\}$ exists, since $1+3=4\neq6$ , $1+4=5\neq 6$ , $1+4+4=9\neq6$ .) Partial converses do exist under extra hypotheses — Cauchy's Theorem (a subgroup of prime order $p$ exists whenever $p \mid |G|$ ) and the Sylow Theorems (subgroups of prime-power order $p^k$ exist whenever $p^k \mid |G|$ ) — but these require separate, deeper arguments and are not consequences of Lagrange's Theorem alone.

The mathematician behind it

Origins · 1 minJoseph-Louis Lagrange and how this was discovered

Joseph-Louis Lagrange

1736–1813 · Turin, Berlin, Paris

Lagrange was one of the towering mathematicians of the 18th century, making foundational contributions to analysis, number theory, mechanics, and the theory of equations. He never stated "Lagrange's Theorem" in the group-theoretic language used today — the abstract notion of a group did not yet exist in his lifetime. His relevant result, from an 1771 memoir on the theory of equations, concerned permutations of the roots of a polynomial and the number of values taken by a rational function under those permutations; in modern language this is a special case (for the symmetric group $S_n$ acting on functions of $n$ variables) of the coset-counting argument now bearing his name. The full abstract statement, for arbitrary finite groups, was crystallized later by mathematicians developing group theory in the 19th century, notably in the wake of Galois's work and subsequent systematizations by Cauchy, Cayley, and others; the name "Lagrange's Theorem" was attached in recognition of Lagrange's priority on the underlying counting idea.

History

- 1770–1771: Lagrange studies permutations of roots of polynomials in his work on solvability of equations, proving a counting result equivalent to a special case of the modern theorem.

- 1799: Ruffini extends permutation-counting ideas in his attempted proof of the unsolvability of the general quintic, implicitly using coset-type counting.

- 1830s: Galois develops the theory of groups of permutations in the context of solvability by radicals, using subgroup counting arguments closely related to Lagrange's original insight.

- 1854: Cayley gives the first abstract definition of a finite group independent of permutations, setting the stage for the fully general statement of the theorem.

- Late 19th century: The theorem is stated and proved in its modern, fully abstract form for arbitrary finite groups, becoming a standard cornerstone of group theory textbooks.

Are left cosets and right cosets always the same set?

Not in general. For a subgroup $H \le G$ , the left coset $gH$ and right coset $Hg$ can differ when $H$ is not a normal subgroup, as shown explicitly in Example 1 with $G = S_3$ . Both collections — the set of all left cosets and the set of all right cosets — still each partition $G$ into pieces of size $|H|$ , and there are always equally many of each type (the map $gH \mapsto Hg^{-1}$ is a bijection between the two collections), so the index $[G:H]$ is the same whether computed via left or right cosets. But the individual cosets themselves need not coincide unless $H$ is normal, in which case $gH = Hg$ for every $g$ , a property developed further under Normal Subgroups.

Does Lagrange's Theorem tell us a subgroup of every divisor order exists?

No — this is a very common misreading. Lagrange's Theorem is a one-directional statement: it says that if $H$ is a subgroup, its order divides $|G|$ . It does not assert the converse, that every divisor of $|G|$ is realized as the order of some subgroup. The counterexample $A_4$ (order 12, no subgroup of order 6) shows the converse fails in general. Partial converses — Cauchy's Theorem for prime divisors, and the Sylow Theorems for prime-power divisors — do hold, but they require substantially more machinery than the coset-counting argument used here.

How does Lagrange's Theorem imply Fermat's Little Theorem?

Let $p$ be prime and $a$ an integer not divisible by $p$ . The set $G = (\mathbb{Z}/p\mathbb{Z})^\times = \{1, 2, \dots, p-1\}$ under multiplication mod $p$ is a group of order $p - 1$ (every nonzero residue mod a prime has a multiplicative inverse). The residue class of $a$ is an element of this group, so by the corollary to Lagrange's Theorem, its order divides $|G| = p-1$ , and hence $a^{p-1} \equiv 1 \pmod p$ . This is exactly Fermat's Little Theorem, recovered as a direct specialization of the "element order divides group order" corollary.

Why must every coset have exactly

|H|

elements — could some cosets be smaller?

No: the bijection $\varphi_g: H \to gH$ given by $\varphi_g(h) = gh$ is always a genuine bijection, for every $g \in G$ , regardless of which coset we look at. Injectivity follows from cancelling $g$ (multiplying by $g^{-1}$ ), and surjectivity onto $gH$ is immediate from the definition of $gH$ as the image of $H$ under left-multiplication by $g$ . Since this argument uses only that $G$ is a group (so $g^{-1}$ exists) and never any special property of $g$ , it applies uniformly to every coset, so all cosets — including $H$ itself, the coset of $e$ — have exactly $|H|$ elements, with no exceptions.

Is it true that

H

itself is always one of the cosets?

Yes. Since $H$ is a subgroup, $e \in H$ , and the coset of the identity element is

eH = \{eh : h \in H\} = \{h : h \in H\} = H.

So $H$ is always exactly one of its own left cosets — namely the coset corresponding to the representative $e$ (or, equivalently, to any representative $g \in H$ , since $gH = H$ whenever $g \in H$ , by the equal-or-disjoint dichotomy applied to $gH \cap H \ni g$ ). This is why the index $[G:H]$ is always at least 1, and equals 1 precisely when $H = G$ .

Does this generalize beyond groups, e.g. to rings or vector spaces?

Yes, the coset idea is far more general than groups. In a vector space $V$ with subspace $W$ , the cosets $v + W$ partition $V$ and form the quotient vector space $V/W$ ; in a ring $R$ with ideal $I$ , the cosets $r + I$ partition $R$ and form the quotient ring $R/I$ . In every one of these settings the underlying mechanism is the same equivalence-relation-and-translation argument used here for groups. The group-theoretic version is the most primitive instance, and the requirement that cosets combine consistently into a well-defined quotient group structure (rather than merely a partition) is exactly what forces the extra hypothesis of normality, explored next under Normal Subgroups.