Subgroups and Lagrange's Theorem

A group within a group

A subgroup is a subset H of a group G that is itself a group under the *same* operation. It isn't enough for H to be a smaller pile of elements; H has to be self-sufficient — closed under the operation, containing the identity, and containing the inverse of each of its members. The even integers form a subgroup of the integers under addition: add two evens and you get an even, 0 is there, and the negative of an even is even.

Every group has at least two subgroups: the whole group itself, and the one-element set {e}. Those are the trivial subgroups; the interesting ones live in between. A reliable way to build one: take any element a and collect all its powers a, a², a³, … together with e and the inverses. That set, written ⟨a⟩, is always a subgroup — the [[cyclic-group|cyclic]] subgroup generated by a — and its size is exactly the order of a.

Lagrange's theorem

Here is the payoff. The number of elements in a group is its order, written |G|. [[lagranges-theorem|Lagrange's theorem]] says: if H is a subgroup of a finite group G, then |H| [[divisibility|divides]] |G|. A group of order 12 can only have subgroups of order 1, 2, 3, 4, 6, or 12 — never 5, never 7. The forbidden sizes are forbidden absolutely.

The idea behind the proof is elegant. Take a subgroup H and “slide” it around the group by picking some element g and forming the set gH = { g ∗ h : h in H }. These slid copies are called cosets. Two facts do all the work: every coset has exactly |H| elements (sliding doesn't shrink or stretch), and any two cosets are either identical or completely disjoint. So the cosets tile G into equal-sized, non-overlapping blocks — and if you can cut a set of size |G| into blocks of size |H|, then |H| must divide |G|.

G = (Z/6Z, ⊕), |G| = 6, identity 0
Subgroup H = {0, 2, 4}  (the multiples of 2 mod 6), |H| = 3.

Check H is a subgroup:  closed? 2⊕2=4 ✓  2⊕4=0 ✓  4⊕4=2 ✓
                        identity 0 ∈ H ✓   inverses: 2↔4, 0↔0 ✓

Cosets of H (slide H by each g = g ⊕ H):
  0 ⊕ H = {0,2,4}        }
  2 ⊕ H = {2,4,0}        }  all equal to H itself
  4 ⊕ H = {4,0,2}        }
  1 ⊕ H = {1,3,5}        }
  3 ⊕ H = {3,5,1}        }  all equal to {1,3,5}
  5 ⊕ H = {5,1,3}        }

Exactly TWO distinct cosets:  {0,2,4} and {1,3,5}.
They tile G:   2 cosets × 3 elements each = 6 = |G|.
So |H| = 3 divides |G| = 6, with index 2.   ✓ Lagrange

Consequences worth keeping

Lagrange pays dividends immediately. Because ⟨a⟩ is a subgroup of size equal to the order of a, the order of every element divides |G| — exactly the pattern you saw in the mod-6 table, where every order was 1, 2, 3, or 6. Two clean corollaries follow:

1. If |G| is a prime number p, the only possible subgroup orders are 1 and p, so any non-identity element generates everything: every group of prime order is [[cyclic-group|cyclic]].
2. Combining a with itself |G| times always returns the identity: a^|G| = e for every a, since the order of a divides |G|.