Lagrange's theorem & cosets

"How big can a subgroup be?" is one of the most basic structural questions about G. Lagrange's theorem nails it down: the order of every subgroup must divide the order of the whole group. A single divisibility constraint that severely restricts what subgroups can exist.

19.1 Proof sketch

Define a ∼ b ⇔ a⁻¹b ∈ H. This is an equivalence relation on G; its classes are the left cosets $gH$, so G partitions into disjoint classes. Each class has size |H| because the map $h\mapsto gh$ is a bijection H → gH. Hence total = (# classes) × (size per class). ∎

19.2 Interactive: pick a subgroup, see its cosets

19.3 Key corollaries

• Element order divides group order: since $⟨g⟩$ is a subgroup of order = ord(g), Lagrange gives $\mathrm{ord}(g)\mid |G|$.
• The max element order on the cube is 1260: because |G| = 2²⁷ · 3¹⁴ · 5³ · 7² · 11, every element order must divide |G|; subject to that constraint, 1260 = 2² · 3² · 5 · 7 is the largest constructible order.
• Groups of prime order are cyclic: if |G| = p (prime), the only subgroups are {e} and G; so any non-identity g generates G. Hence G ≅ ℤ/p.

19.4 Cauchy's theorem — partial converse to Lagrange

For the cube, the prime divisors of |G| are $\{2,3,5,7,11\}$. Cauchy guarantees that G contains elements of order exactly 2, 3, 5, 7, 11:

19.5 Sylow theorems — Cauchy's full strengthening

Lagrange gives only necessity; Cauchy provides existence at prime order; Sylow's theorems (1872) precisely describe all prime-power-order subgroups. Write $|G|=p^a\cdot m$ with $\gcd (p,m)=1$.

For the cube, |G| = 2²⁷ · 3¹⁴ · 5³ · 7² · 11. Sylow subgroup orders are: