Abstract Algebra

Finite abelian group structure

The fundamental theorem of finite abelian groups states that every finite abelian group is a direct product of cyclic groups of prime-power order, and this decomposition is unique up to the order of the factors. If |G| = \prod_i p_i^{k_i}, then G is the direct product of its Sylow p_i-subgroups, and each Sylow component is a direct product of cyclic p_i-groups. So the number of isomorphism classes of abelian groups of order n equals the product, over the primes p dividing n, of the number of partitions of the exponent of p. For example, the abelian groups of order 2^3\cdot 3^2 number 3\times 2=6, because 3 has three partitions and 2 has two.

A cyclic group of order n has exactly one subgroup of each order d with d\mid n, so its number of subgroups equals the number of divisors of n. For instance, a cyclic group of order 8 has four subgroups, of orders 1,2,4,8.

For a direct product \mathbb{Z}_m\oplus\mathbb{Z}_n, the order of an element (a,b) is the least common multiple of the orders of a in \mathbb{Z}_m and b in \mathbb{Z}_n, and the order of a in \mathbb{Z}_m is m/\gcd(a,m). The exponent of the group is the lcm of m and n. A finite abelian group is cyclic exactly when its order equals its exponent, and \mathbb{Z}_m\oplus\mathbb{Z}_n is cyclic exactly when \gcd(m,n)=1. In particular, a group whose order exceeds its exponent has no element of order equal to its order.

The multiplicative group F^\times of a finite field with q elements is cyclic of order q-1. Hence it has a subgroup of order d exactly when d\mid q-1, which is the criterion used to find the smallest admissible field order for a given subgroup order.

Sylow theorems

Let |G| = p^a m with p\nmid m. The first Sylow theorem guarantees a subgroup of order p^a, called a Sylow p-subgroup. The second theorem says all Sylow p-subgroups are conjugate, and the third theorem restricts their number n_p by n_p\equiv 1 \pmod p and n_p\mid m. A Sylow p-subgroup is normal if and only if n_p=1, and in that case it is the unique Sylow p-subgroup.

In practice, the Sylow conditions are often combined for several primes. For each prime one lists the divisors of the cofactor m that are congruent to 1 modulo p, and if only 1 survives, the Sylow subgroup is forced to be normal. This method determines the possible values of n_p for any group order, and it can show that n_{11} and n_{23} are restricted in a group of order 2024. For example, the number of Sylow 5-subgroups of S_5 is 6, because each such subgroup is generated by a 5-cycle, each contains four 5-cycles, and there are 24 five-cycles in total.

For a group of order 595 = 5\cdot 7\cdot 17, the divisors of 119 congruent to 1 modulo 5 give only n_5=1, so the Sylow 5-subgroup is normal and proper. Cauchy's theorem also guarantees an element of order 17 in this group, since 17 divides the order.

Quotient rings and units

For an ideal I of a commutative ring R, the quotient ring R/I has cosets g+I as elements, and g+I is a unit exactly when there exists h with gh-1\in I. In F[x] over a field F, a coset g+I with I=\langle f\rangle is a unit precisely when \gcd(g,f)=1. This makes it possible to count units in a finite quotient ring by counting the residues coprime to the modulus.

For example, in \mathbb{Z}_3[x]/\langle (x-1)^2\rangle there are nine cosets, each represented uniquely as a+bx with a,b\in\mathbb{Z}_3. A coset is a unit exactly when its value at x=1 is nonzero, that is, when a+b\not\equiv 0 \pmod 3, and this gives six units.

The ring \mathbb{R}[X]/\langle X^4+4\rangle is not a field and not an integral domain. Factoring X^4+4=(X^2+2X+2)(X^2-2X+2) into two coprime irreducible quadratics and applying the Chinese Remainder Theorem gives a product of two copies of \mathbb{C}. A product of two nonzero rings has zero divisors, but its only nilpotent element is 0, since an element (a,b) with a^k=b^k=0 in \mathbb{C} must be zero.

A local ring has a unique maximal ideal, and every element outside it is a unit. In \mathbb{Z}[x]/\langle 2,x^3\rangle\cong \mathbb{F}_2[x]/\langle x^3\rangle, the maximal ideal \langle x\rangle is nilpotent, so every element is either a unit or nilpotent. The ring \mathbb{Z}_2[x]/\langle x^2\rangle has the same behaviour with x^2=0, and it has exactly one maximal ideal.

Class equation of finite groups

A group acts on itself by conjugation, and the orbits of this action are the conjugacy classes. The size of the class of x equals the index [G:C_G(x)] of its centralizer, so it divides |G|. The class equation states that |G| = |Z(G)| + \sum (sizes of the non-central classes), and the number of conjugacy classes is the number of orbits. For example, S_3 has order 6 = 1+2+3, so it has three classes, namely the identity, the three transpositions, and the two 3-cycles.

For a p-group, every centralizer has p-power index in the group, so every class size is a power of p, and the center is nontrivial. If every non-central element has centralizer of order p^3 in a group of order p^4, then each non-central class has p elements, and the total number of classes equals |Z(G)|+(|G|-|Z(G)|)/p. This is the kind of count the class equation is used to compute.

An abelian group has as many conjugacy classes as elements, because every class is a singleton. Consequently the only abelian group with exactly three conjugacy classes is the cyclic group of order 3. Among non-abelian groups, the group with exactly three classes is S_3, so there are two non-isomorphic groups with exactly three conjugacy classes.

Comaximal ideals and Chinese remainder theorem

Two ideals J_1 and J_2 of a commutative ring R are called comaximal when J_1+J_2=R. If they are comaximal, then J_1\cap J_2 = J_1J_2 and the Chinese Remainder Theorem gives R/(J_1\cap J_2)\cong R/J_1\times R/J_2. For example, 2\mathbb{Z}+3\mathbb{Z}=\mathbb{Z}, so \mathbb{Z}/6\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}.

The condition that every pair of residues is achieved by a unique element is stronger than comaximality. Existence for all pairs forces J_1+J_2=R, while uniqueness forces J_1\cap J_2=0, because two solutions differ by an element of the intersection. So existence and uniqueness together imply comaximality, but comaximality alone does not give uniqueness. In \mathbb{Z}, the ideals 2\mathbb{Z} and 3\mathbb{Z} are comaximal, yet their intersection 6\mathbb{Z} is nonzero, so uniqueness fails.

For a commutative ring with 1\neq 0, the following are equivalent to the existence of a nontrivial idempotent e with e\neq 0,1: R is a product of two nonzero rings, R is the sum of two nonzero proper ideals with zero intersection, and R has two elements a,b with a\neq 0\neq b and ab=0 are weaker. A nontrivial idempotent e gives R\cong Re\times R(1-e), and the ideals Re and R(1-e) are comaximal with zero intersection. The converse of the zero-divisor statement fails: in \mathbb{Z}/4\mathbb{Z} the element 2 satisfies 2\cdot 2=0, but the only idempotents are 0 and 1, so no product decomposition exists.

Minimal polynomials and field extensions

If \alpha is algebraic over a field F, its minimal polynomial p(x) is the monic irreducible polynomial of least degree over F having \alpha as a root. Then F(\alpha)\cong F[x]/\langle p\rangle and [F(\alpha):F]=\deg p. The tower law [K:F]=[K:L][L:F] is the basic tool for comparing degrees across a chain of fields.

When passing to a smaller field L\subseteq F that contains all coefficients of p, the polynomial p stays irreducible over L, because any factorization over L is also a factorization over F. Therefore [L(\alpha):L]=[F(\alpha):F]. Applying the tower law to F(\alpha)\supseteq L(\alpha)\supseteq L gives [F(\alpha):L(\alpha)]=[F:L]. Consequently F(\alpha) is a finite extension of L(\alpha) exactly when F is a finite extension of L, and the same equivalence holds for algebraicity.

A polynomial f over a finite field \mathbb{Z}_p gives a field \mathbb{Z}_p[x]/\langle f\rangle exactly when f is irreducible, and then its degree over \mathbb{Z}_p equals \deg f. Over \mathbb{Z}_3, a cubic is irreducible precisely when it has no root in \mathbb{Z}_3, so one checks the three values at 0,1,2. When no value vanishes, the quotient is a field with 3^3 elements, and its degree is 3.

Polynomial rings over fields

For a field F, the polynomial ring F[x] is a principal ideal domain, and therefore a unique factorization domain. In a PID, a nonzero ideal generated by an irreducible element is maximal, and every maximal ideal is prime. Since the converse of the last statement fails in general, one must check each property separately. Over a field, the quotient F[x]/\langle f\rangle is a field exactly when f is irreducible.

Over \mathbb{Z}_p with p odd, a quadratic x^2+bx+c is irreducible exactly when its discriminant b^2-4c is a non-square modulo p, since then it has no root. For example, if the discriminant is a non-square modulo 11, the quadratic is irreducible, the quotient by the ideal it generates is a field, and the ideal is maximal and therefore prime.

A unique factorization domain must be an integral domain, so any ring with zero divisors fails to be a UFD. The ring of 2\times 2 real matrices is non-commutative and has zero divisors, and so does the ring of all real-valued functions on a set, where two functions with disjoint nonzero supports multiply to zero. The ring of Riemann integrable functions on [0,1] also has zero divisors, since functions supported on disjoint subintervals multiply to zero. By contrast, \mathbb{R}[x] is a PID and hence a UFD.