Lagrange's Group Theorem — Definition, Formula & Examples

Lagrange's Group Theorem states that for any finite group, the number of elements in a subgroup always divides evenly into the number of elements in the whole group. This means a group of 12 elements can have subgroups of order 1, 2, 3, 4, 6, or 12, but never of order 5 or 7.

Let $G$ be a finite group and $H$ a subgroup of $G$ . Then the order of $H$ divides the order of $G$ , and the quotient $|G|/|H|$ equals the number of distinct left cosets of $H$ in $G$ , called the index of $H$ in $G$ , denoted $[G:H]$ .

Key Formula

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

Where:

$|G|$ = Order (number of elements) of the finite group G
$|H|$ = Order of the subgroup H
$[G:H]$ = Index of H in G — the number of distinct left cosets of H

How It Works

The proof relies on partitioning

G

into left cosets of

H

. Each left coset

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

has exactly

|H|

elements, and distinct cosets are disjoint. Since these cosets cover all of

G

, the total number of elements satisfies

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

, which immediately gives the divisibility result. A key corollary is that the order of any element

g \in G

divides

|G|

, since the order of

g

equals the order of the cyclic subgroup

\langle g \rangle

Worked Example

Problem: Let G = S₃, the symmetric group on 3 elements, which has order 6. Verify Lagrange's Theorem for the subgroup H = {e, (1 2)} generated by the transposition (1 2).

Step 1: Determine the order of each set. S₃ has all permutations of {1, 2, 3}, so |G| = 6. The subgroup H = {e, (1 2)} has |H| = 2.

|G| = 6, \quad |H| = 2

Step 2: Check divisibility. Since 2 divides 6, Lagrange's Theorem is satisfied.

\frac{|G|}{|H|} = \frac{6}{2} = 3

Step 3: Verify by listing the 3 left cosets: eH = {e, (1 2)}, (1 3)H = {(1 3), (1 2 3)}, and (2 3)H = {(2 3), (1 3 2)}. These are disjoint and cover all of S₃.

S_3 = eH \cup (1\;3)H \cup (2\;3)H

Answer: |H| = 2 divides |G| = 6, with index [G : H] = 3, confirming Lagrange's Theorem.

Why It Matters

Lagrange's Theorem is the first major constraint on subgroup structure and underpins results like Fermat's Little Theorem in number theory and Cauchy's Theorem in group theory. It appears constantly in abstract algebra courses and is essential for classifying finite groups.

Common Mistakes

Mistake: Assuming the converse: if d divides |G|, then G must have a subgroup of order d.

Correction: The converse of Lagrange's Theorem is false in general. For example, the alternating group A₄ has order 12 but no subgroup of order 6. A partial converse holds by Cauchy's Theorem (for prime divisors) and Sylow's Theorems (for prime-power divisors).