TheoremProved

Lagrange's theorem (group theory)

Statement

If $G$ is a finite group and $H$ is a subgroup of $G$ , then $|H|$ divides $|G|$ ; in fact $|G| = [G:H]\cdot|H|$ , where $[G:H]$ is the number of cosets of $H$ in $G$ .

Why is it true?

A subgroup $H$ slices the whole group $G$ into cosets $gH$ that are all exactly the same size as $H$ (translating by $g$ is a bijection) and that never overlap — every element of $G$ lands in exactly one coset. Since $G$ is completely tiled by same-sized, non-overlapping pieces, its size must be a whole multiple of the piece size $|H|$ .

Proof sketch

Show that the left cosets $gH$ partition $G$ : they cover $G$ (each $g\in gH$ ) and any two cosets are either identical or disjoint (if $g_1H\cap g_2H\neq\varnothing$ then $g_1H=g_2H$ ). Each coset has exactly $|H|$ elements because $h\mapsto gh$ is a bijection $H\to gH$ . Summing $|H|$ over the $[G:H]$ cosets gives $|G|=[G:H]\cdot|H|$ .

Stated by

stated 1771Joseph-Louis Lagrange

Topics that use this theorem

Group theory

Related theorems

Fundamental theorem of Galois theory

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

David S. Dummit, Richard M. Foote (2004). Abstract Algebra