Breaking a group into simple pieces
A composition series is a chain 1 = G₀ ◁ G₁ ◁ ··· ◁ Gₙ = G in which each quotient Gᵢ₊₁/Gᵢ is a simple group — you cannot refine the chain any further. The Jordan-Holder theorem says any two composition series of the same finite group have the same length and the same multiset of simple quotients (the composition factors), up to reordering and isomorphism. These factors are the “primes” of the group: like the prime factorization of an integer, they are an invariant, though they do not by themselves reconstruct G (different groups can share composition factors). The classification of finite simple groups names every possible factor.
Two composition series of Z/12Z, same factors (Jordan-Holder).
Series A: 0 ◁ Z/2 ◁ Z/6 ◁ Z/12
factors: (Z/2)/0 = Z/2, (Z/6)/(Z/2) = Z/3, (Z/12)/(Z/6) = Z/2
multiset of factors: { Z/2, Z/2, Z/3 }
Series B: 0 ◁ Z/3 ◁ Z/6 ◁ Z/12
factors: Z/3, Z/2, Z/2
multiset of factors: { Z/2, Z/2, Z/3 }
Same multiset {Z/2, Z/2, Z/3}, just reordered — exactly as Jordan-Holder
predicts. Note 12 = 2 * 2 * 3: the composition factors of a finite
abelian group recover its prime factorization.Solvable and nilpotent groups
A finite group is solvable exactly when all its composition factors are abelian — equivalently, when the derived series G ▷ G′ ▷ G″ ▷ ··· (iterating the commutator subgroup) reaches 1. Solvability is the group-theoretic shadow of solvability by radicals: a polynomial is solvable by radicals iff its Galois group is a solvable group, which is precisely why the general quintic is not solvable — S₅ is not, because A₅ (a non-abelian simple group of order 60) appears as a composition factor.
A nilpotent group is a stronger condition: its upper central series 1 ◁ Z(G) ◁ Z₂(G) ◁ ··· climbs all the way to G. Every nilpotent group is solvable but not conversely (S₃ is solvable, not nilpotent). The clean structural fact: a finite group is nilpotent iff it is the direct product of its Sylow subgroups — so finite nilpotent groups are, in a precise sense, just bundles of p-groups. This is where guides 2 and 4 reconverge.
Presentations and the structure of f.g. abelian groups
How do you write down a group concretely? A presentation ⟨ S | R ⟩ specifies generators S and relations R: it is the free group on S — the group of reduced words with no relations imposed — modulo the smallest normal subgroup containing R. For example ⟨ r, s | rⁿ = 1, s² = 1, srs = r⁻¹ ⟩ is the dihedral group of order 2n. Presentations are how infinite groups, and groups defined by topology, get pinned down; they are also where genuine undecidability lives (the word problem can be unsolvable).
For abelian groups everything becomes computable. The classification of finitely generated abelian groups says every such group is isomorphic to Zʳ ⊕ Z/d₁Z ⊕ ··· ⊕ Z/d_kZ with d₁ | d₂ | ··· | d_k (invariant factor form), or equivalently to Zʳ ⊕ (a product of Z/pⁱZ's) (primary form). The free rank r and the divisors dᵢ are complete invariants. This is the abelian special case of the structure theorem for modules over a PID: present the group as cokernel of an integer matrix, then run Smith normal form to read off the dᵢ.
Classify the abelian group A = <x, y | 6x + 2y = 0, 2x + 4y = 0> (written additively). Relation matrix M (rows = relations, cols = generators x, y): M = [6, 2; 2, 4] Smith normal form via integer row/column operations: [6, 2; 2, 4] swap rows -> [2, 4; 6, 2] R2 <- R2 - 3 R1 -> [2, 4; 0, -10] C2 <- C2 - 2 C1 -> [2, 0; 0, -10] -> diag(2, 10) Invariant factors: d_1 = 2, d_2 = 10, and 2 | 10. Free rank r = (#gens) - (#nonzero d_i) = 0. Therefore A ≅ Z/2Z ⊕ Z/10Z. Primary form: 10 = 2 * 5, so A ≅ Z/2Z ⊕ Z/2Z ⊕ Z/5Z. Order of A = 2 * 10 = 20, matching |det M| = |6*4 - 2*2| = 20.