Why A_n is simple for n ≥ 5 and what that means for polynomials
Every permutation of n objects can be written as a product of transpositions — swaps of exactly two elements. While this decomposition is not unique, the parity of the number of transpositions is: a permutation is always expressed as an even number of transpositions or always as an odd number. This fundamental theorem allows us to assign a well-defined sign (or signature) to each permutation: sgn(σ) = +1 for even permutations and sgn(σ) = −1 for odd permutations.
The sign function sgn: Sₙ → {±1} is a group homomorphism from the symmetric group Sₙ to the multiplicative group {+1, −1}. The kernel of this homomorphism — the set of all even permutations — forms a normal subgroup of Sₙ called the alternating group Aₙ. Since half of all permutations are even and half are odd, |Aₙ| = n!/2. The alternating group is the unique subgroup of index 2 in Sₙ (for n ≥ 2), and it plays a starring role in the classification of finite simple groups.
For small n, the alternating groups are familiar objects. A₁ and A₂ are trivial (one element). A₃ ≅ Z/3Z is cyclic of prime order. A₄ has 12 elements and is not simple — it contains the Klein four-group V₄ = {e, (12)(34), (13)(24), (14)(23)} as a normal subgroup. The drama begins at n = 5.
The alternating group A₅ has |A₅| = 60 elements. To show it is simple, we must show it has no normal subgroups of order d where 1 < d < 60 and d divides 60. The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60, so we must rule out normal subgroups of each intermediate order. The key tool is the class equation: the conjugacy classes of A₅ have sizes 1, 12, 12, 15, and 20.
Any normal subgroup N of A₅ must be a union of conjugacy classes (since normality means N is closed under conjugation) and must contain the identity (a class of size 1). So |N| must equal 1 plus some sum of elements from {12, 12, 15, 20}. One checks that no such sum yields a proper divisor of 60 greater than 1: 1 + 12 = 13, 1 + 15 = 16, 1 + 20 = 21, 1 + 12 + 12 = 25, 1 + 12 + 15 = 28, and so on — none of these divide 60. Therefore A₅ has no proper non-trivial normal subgroups and is simple.
This proof illustrates a technique that recurs throughout the classification program: analyzing groups through their conjugacy classes and using divisibility constraints from Lagrange's theorem. The simplicity of A₅ was first established by Galois around 1830, and it was the key insight behind one of the most famous impossibility results in mathematics: the unsolvability of the general quintic equation by radicals.
One of the most celebrated results in mathematics is that there is no general formula — in terms of radicals (nth roots, additions, and multiplications) — for the solutions of polynomial equations of degree 5 or higher. The quadratic formula, cubic formula, and quartic formula exist, but the quintic has no such counterpart. The proof, due to Galois and Abel, hinges on the simplicity of A₅.
Galois showed that a polynomial equation is solvable by radicals if and only if its Galois group is a solvable group — a group whose composition factors are all cyclic of prime order. The general quintic has Galois group S₅, which has composition factors Z/2Z and A₅. Since A₅ is simple but not cyclic of prime order (it is non-abelian with 60 elements), S₅ is not solvable, and the general quintic is not solvable by radicals. This was the birth of group theory as a discipline: Galois invented the subject precisely to answer this question.
The connection runs even deeper. A group is solvable if and only if all its composition factors are abelian simple groups (i.e., cyclic of prime order). The existence of a non-abelian simple group — A₅ — is what creates the obstruction. In a precise sense, A₅ is the reason the quintic cannot be solved. This makes A₅ one of the most consequential mathematical objects ever discovered, and it was the first hint that simple groups would be central to understanding the structure of mathematics.
The simplicity of A₅ is not an isolated phenomenon — it extends to all alternating groups of degree 5 or greater. The standard proof proceeds by showing that every normal subgroup of Aₙ (for n ≥ 5) must contain a 3-cycle, and then showing that any normal subgroup containing a 3-cycle must contain all 3-cycles, hence must be all of Aₙ. The key lemma is that any two 3-cycles in Aₙ are conjugate when n ≥ 5, which fails for n = 4 (explaining why A₄ is not simple).
The alternating groups thus form the first infinite family of non-abelian simple groups. Their orders |Aₙ| = n!/2 grow rapidly: |A₅| = 60, |A₆| = 360, |A₇| = 2520, and so on. These groups are intimately connected to combinatorics, geometry, and the theory of equations. For example, A₅ is isomorphic to PSL(2, 4) ≅ PSL(2, 5), providing a bridge between the alternating groups and the groups of Lie type that we will encounter in the next lesson.
A natural question is whether the alternating groups are the "smallest" non-abelian simple groups. In a sense, yes: A₅ with its 60 elements is the smallest non-abelian simple group. But the landscape quickly becomes rich and complex. The next non-abelian simple group by order is PSL(2, 7), with 168 elements, which is a group of Lie type. The interleaving of alternating groups and groups of Lie type in the ordering by size is one of the beautiful patterns revealed by the classification.