Subgroups, Cosets, and Lagrange's Theorem

⚡ Key Takeaways

Subgroups, Cosets, and Lagrange's Theorem. A subgroup is a subset of a group that is itself a group under the inherited operation; cosets are the translates of a subgroup that partition the whole group into equal-sized pieces; Lagrange's Theorem states that the order of any subgroup of a finite group divides the order of the group. Applies throughout finite group theory, number theory (Fermat/Euler), and the classification of small groups. Named for Joseph-Louis Lagrange (1770s), formalized in the 19th century.

💡 The Core Insight

A subgroup acts on its parent group by translation, and translation is a bijection. Therefore every coset has exactly as many elements as the subgroup, and the cosets tile the group without overlap. Counting tiles times tile-size gives the group's order — so the subgroup's order must divide it. Almost everything about finite groups starts from this counting argument.

DefinitionSubgroup

Let $(G,\cdot)$ be a group with identity $e$ . A subset $H \subseteq G$ is a subgroup, written $H \le G$ , if:

1

\quad H \neq \varnothing, \qquad \forall a,b \in H:\ ab \in H, \qquad \forall a \in H:\ a^{-1} \in H.

Derivation hint: These three conditions (nonempty, closed under product, closed under inverse) guarantee $(H,\cdot)$ satisfies all group axioms, since associativity is inherited and $e = aa^{-1} \in H$ . Equivalently (the one-step subgroup test), a nonempty subset $H$ is a subgroup iff:

2

\quad \forall a,b \in H:\ ab^{-1} \in H.

Derivation hint: Taking $a=b$ gives $e \in H$ ; taking $a=e$ gives $b^{-1} \in H$ ; then $a(b^{-1})^{-1} = ab \in H$ recovers closure.

DefinitionCosets and Index

Let $H \le G$ and $g \in G$ . The left coset of $H$ by $g$ and the right coset are:

3

\quad gH = \{\, gh : h \in H \,\}, \qquad Hg = \{\, hg : h \in H \,\}.

Derivation hint: A coset is the image of $H$ under left- (or right-) multiplication by the fixed element $g$ . The index $[G:H]$ is the number of distinct left cosets of $H$ in $G$ :

4

\quad [G:H] = \bigl|\, \{\, gH : g \in G \,\} \,\bigr| = |G/H|.

Derivation hint: $G/H$ denotes the set of left cosets; the index counts the blocks of the coset partition.

TheoremLagrange's Theorem

If $G$ is a finite group and $H \le G$ , then

5

\quad |G| = [G:H]\,\cdot\,|H|, \qquad \text{hence } |H| \ \big|\ |G|.

Derivation hint: The $[G:H]$ left cosets partition $G$ , and each has exactly $|H|$ elements, so total size = (number of cosets) × (size of each coset). In particular, for every $g \in G$ the order of $g$ (the least $n>0$ with $g^n = e$ ) divides $|G|$ , since $\langle g \rangle \le G$ has order equal to $\operatorname{ord}(g)$ :

6

\quad \operatorname{ord}(g) \ \big|\ |G| \qquad \text{and} \qquad g^{|G|} = e.

Derivation hint: Apply [5] to the cyclic subgroup $\langle g \rangle$ ; then $|G| = k\cdot\operatorname{ord}(g)$ gives $g^{|G|} = (g^{\operatorname{ord}(g)})^k = e$ .

ProofCosets partition

G

into equal-sized blocks

We prove Lagrange's Theorem by establishing two lemmas, then counting.

Step 1 — Define an equivalence relation. For $a,b \in G$ declare $a \sim b$ iff $a^{-1}b \in H$ . This is:

reflexive: $a^{-1}a = e \in H$ ;
symmetric: if $a^{-1}b \in H$ then $(a^{-1}b)^{-1} = b^{-1}a \in H$ ;
transitive: if $a^{-1}b, b^{-1}c \in H$ then $(a^{-1}b)(b^{-1}c) = a^{-1}c \in H$ .

Step 2 — Equivalence classes are left cosets. The class of $a$ is

7

\quad [a] = \{\, b \in G : a^{-1}b \in H \,\} = \{\, b : b = ah,\ h \in H \,\} = aH.

Derivation hint: $a^{-1}b \in H \iff b \in aH$ , by left-multiplying the membership condition by $a$ . Since the classes of an equivalence relation partition the set, the left cosets partition $G$ : distinct cosets are disjoint and their union is all of $G$ .

Step 3 — Every coset has size $|H|$ . Fix $g$ . The map $\varphi : H \to gH$ , $\varphi(h) = gh$ , is:

surjective by definition of $gH$ ;
injective because $gh_1 = gh_2 \Rightarrow h_1 = h_2$ by left-cancellation.

8

\quad |gH| = |H| \quad\text{for every } g \in G.

Derivation hint: Left multiplication by $g$ is a bijection $G \to G$ (inverse: multiply by $g^{-1}$ ), so it bijects $H$ onto $gH$ .

Step 4 — Count. Let the distinct cosets be $g_1H, \dots, g_kH$ where $k = [G:H]$ . By Steps 2–3,

9

\quad |G| = \sum_{i=1}^{k} |g_iH| = \sum_{i=1}^{k} |H| = k\,|H| = [G:H]\,|H|.

Derivation hint: Disjoint union (Step 2) lets sizes add; each summand equals $|H|$ (Step 3). Therefore $|H|$ divides $|G|$ . $\blacksquare$

∎

⚠️ Left vs. right cosets need not coincide

In general $gH \ne Hg$ . They do coincide for every $g$ exactly when $H$ is a normal subgroup. But Lagrange's Theorem holds regardless: the number of left cosets equals the number of right cosets, because $gH \mapsto Hg^{-1}$ is a bijection between the two sets of cosets, so $[G:H]$ is well-defined either way.

mathematician · Joseph-Louis Lagrange | 1736–1813

Lagrange, working in pre-group-theoretic language, studied how the value of a polynomial in $n$ variables changes when its variables are permuted. In his 1770–71 memoir Réflexions sur la résolution algébrique des équations, he observed that the number of distinct values a rational function takes under all $n!$ permutations always divides $n!$ . This is precisely the index formula $[S_n : H]$ applied to the stabilizer $H$ of the function — the embryonic form of "Lagrange's Theorem," decades before Galois, Cauchy, and Cayley gave groups their abstract definition. The theorem now bearing his name is the abstract distillation of this combinatorial observation.

🧠 Intuition

Think of a subgroup $H$ as a "pattern" or "tile" sitting inside $G$ , anchored at the identity. Multiplying everything in $H$ by a fixed element $g$ slides the whole tile to a new location $gH$ without distorting it — translation is rigid, so the slid tile has the same number of points. Now slide the tile to every possible position. Two slid tiles either land on exactly the same spot (if $g_1^{-1}g_2 \in H$ ) or are completely disjoint — there is no partial overlap, because membership in a coset is an all-or-nothing equivalence. So the group floor is tiled perfectly by congruent copies of $H$ . Counting the floor area two ways (total points vs. tiles × points-per-tile) forces $|H|$ to divide $|G|$ . The integers $\bmod n$ make this concrete: $H = \{0,3\} \subseteq \mathbb Z_6$ tiles $\mathbb Z_6$ as $\{0,3\}, \{1,4\}, \{2,5\}$ — three tiles of two, and $3\times 2 = 6$ .

Figure

Cosets of

H=\{0,3\}

tiling

\mathbb{Z}_6

into three equal blocks

ExampleCosets and index in

\mathbb{Z}_{12}

Given. $G = (\mathbb{Z}_{12}, +)$ , $|G| = 12$ . Take $H = \langle 4 \rangle = \{0, 4, 8\}$ .

Step 1 — Verify $H \le G$ . Closed: $4+4=8$ , $4+8=0$ , $8+8=4$ all in $H$ ; inverses: $-4 = 8 \in H$ , $-8 = 4 \in H$ ; $0 \in H$ . So $H \le G$ with $|H| = 3$ .

Step 2 — Compute the cosets (additive, so $g+H$ ).

$0+H = \{0,4,8\}$
$1+H = \{1,5,9\}$
$2+H = \{2,6,10\}$
$3+H = \{3,7,11\}$
$4+H = \{4,8,0\} = 0+H$ (repeat)

Step 3 — Count distinct cosets. There are exactly 4 distinct cosets, each of size 3.

Step 4 — Check Lagrange. $|G| = [G:H]\,|H| = 4 \times 3 = 12.$ ✓

\boxed{\,[G:H] = 4, \quad |H| = 3, \quad 3 \mid 12\,}

ExampleElement orders via Lagrange in

S_3

Given. $G = S_3$ , the symmetric group on $\{1,2,3\}$ , with $|G| = 6$ .

Step 1 — Possible element orders. By [6], $\operatorname{ord}(g) \mid 6$ , so each order $\in \{1,2,3,6\}$ .

Step 2 — Exclude order 6. If some $g$ had order 6, $\langle g \rangle$ would be all of $S_3$ , making $S_3$ cyclic and hence abelian. But $S_3$ is non-abelian (e.g. $(12)(13) = (132)$ while $(13)(12)=(123)$ ). So no element of order 6.

Step 3 — Tabulate.

identity $e$ : order 1.
transpositions $(12),(13),(23)$ : order 2 (each squares to $e$ ).
3-cycles $(123),(132)$ : order 3.

Step 4 — Verify divisibility. Orders observed: $\{1,2,3\}$ , all divide 6. ✓ Subgroup orders that occur: $1, 2, 3, 6$ — all divisors of 6 are realized here.

\boxed{\,\text{Element orders in } S_3 = \{1,2,3\}, \text{ each dividing } |S_3|=6\,}

⚠️ The converse of Lagrange is FALSE

Lagrange's Theorem says: if $H$ is a subgroup then $|H| \mid |G|$ . It does not say that every divisor of $|G|$ is the order of some subgroup. Students routinely assume "divisor $\Rightarrow$ subgroup exists," which is wrong in general. Partial converses do hold:

Cauchy's Theorem: if a prime $p$ divides $|G|$ , there is an element (and subgroup) of order $p$ .
Sylow's Theorems: if $p^k$ is the largest power of $p$ dividing $|G|$ , a subgroup of order $p^k$ exists.
Finite abelian / cyclic groups: the full converse holds — every divisor is achieved.

But for general $G$ , composite-order subgroups can simply fail to exist.

Counterexample

A_4

has no subgroup of order 6

The alternating group $A_4$ has order $12$ , and $6 \mid 12$ , yet $A_4$ has no subgroup of order 6. Suppose $H \le A_4$ with $|H| = 6$ . Then $[A_4 : H] = 2$ , which would force $H$ to be normal, and the quotient $A_4/H \cong \mathbb{Z}_2$ . Counting: $A_4$ contains eight 3-cycles, each of order 3. A subgroup of index 2 must contain the square of every element (since in $A_4/H \cong \mathbb Z_2$ , $g^2 \in H$ always); but for a 3-cycle $g$ , $g = g^4 = (g^2)^2 \in H$ , so all eight 3-cycles lie in $H$ — already more than 6 elements. Contradiction. Hence no order-6 subgroup exists, definitively refuting the converse of Lagrange.

connection

Lagrange's Theorem is the gateway to the structural theory of finite groups. It directly feeds Cyclic Groups (a group of prime order is forced to be cyclic since every non-identity element has order $p$ ) and motivates Normal Subgroups and Quotient Groups, where cosets of a normal subgroup acquire their own group structure $G/N$ . The proof's machinery — translating a subgroup by group elements — is a special case of Group Actions and the Orbit-Stabilizer Theorem, in which $|G| = |\text{orbit}|\cdot|\text{stabilizer}|$ generalizes [5]. Its partial converses are Cauchy's Theorem and Sylow Theorems. In number theory the corollary $g^{|G|}=e$ specializes, via the unit group $(\mathbb Z/n\mathbb Z)^\times$ , to Euler's Theorem and Fermat's Little Theorem. The index also underlies the First Isomorphism Theorem through $|G/\ker\varphi| = |\operatorname{im}\varphi|$ .

🔬 Where this appears

⏳ History

- 1770–1771 — Lagrange's memoir on equation-solving notes that the number of values of a polynomial under variable permutations divides

n!

— the seed of the theorem.

- 1799 — Paolo Ruffini uses permutation-substitution counting (implicitly coset-like) in attempts to prove the unsolvability of the quintic.

- 1815 — Augustin-Louis Cauchy systematically studies "substitutions" (permutations), proving results on cycle structure and laying groundwork for subgroup counting.

- 1830–1832 — Évariste Galois introduces normal subgroups ("invariant subgroups") and proper decompositions, deepening the coset idea in his theory of solvability.

- 1845–1846 — Cauchy proves what is now Cauchy's Theorem (prime divisors yield elements of that order), a partial converse to Lagrange.

- 1854 — Arthur Cayley gives the first abstract definition of a group in On the theory of groups, as depending on the symbolic equation $\theta^n = 1$ , allowing Lagrange's result to be stated for abstract groups.

- 1872 — Ludwig Sylow publishes the Sylow Theorems, the deepest partial converses, guaranteeing subgroups of prime-power order.

- 1870s–1880s — Camille Jordan and Walther von Dyck consolidate abstract group theory; cosets, index, and Lagrange's Theorem take their modern textbook form.

- 1897 — William Burnside's Theory of Groups of Finite Order presents the topic in essentially the form taught today.

open-problem

Where Lagrange stops short. Lagrange's Theorem is a one-directional divisibility constraint; the genuinely hard questions concern which divisors are realized and how many subgroups exist:

Counting subgroups of given order. Determining the number of subgroups of each order in an arbitrary finite group is computationally and theoretically intricate; for $p$ -groups it connects to deep questions about subgroup lattices and Hall's enumeration formulas.
CLT groups. A group satisfying the full converse of Lagrange (a subgroup of every order dividing $|G|$ ) is called a converse-Lagrange-theorem (CLT) group. Characterizing CLT groups is subtle: all supersolvable groups are CLT, but the property is not closed under quotients or subgroups in obvious ways, and a clean intrinsic characterization remains messy.
Infinite groups. For infinite $G$ the equation $|G| = [G:H]\,|H|$ becomes a statement about cardinal arithmetic and generally gives no useful constraint; "order divides order" loses meaning, so Lagrange's combinatorial force is genuinely a finiteness phenomenon.
Beyond groups. Analogues for loops, quasigroups, and Moufang loops are delicate — Lagrange's Theorem can fail for general quasigroups, and its validity for Moufang loops was a long-studied question resolved only with substantial machinery.

Subgroups, Cosets, and Lagrange's Theorem

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also