Modern Algebra

UPSC Mathematics Optional · Paper II · Groups, Rings & Fields

1. Groups — Basics 2. Subgroups & Cosets 3. Normal Subgroups & Quotients 4. Homomorphisms & Isomorphisms 5. Cyclic & Permutation Groups 6. Sylow Theorems 7. Rings & Ideals 8. Integral Domains & Fields 9. Polynomial Rings & Factorization

§1 · Groups — Basics

Group Axioms

A set \(G\) with binary operation \(*\) is a group if: closure, associativity, identity (\(e\)), inverses.

Abelian: also commutative. Order of group \(|G|\). Order of element \(a\) = smallest \(n\) with \(a^n=e\).

Basic Results PYQ★★

Identity is unique. Inverse of each element is unique.
\((ab)^{-1}=b^{-1}a^{-1}\). \((a^n)^{-1}=a^{-n}\).
Cancellation laws hold: \(ab=ac\Rightarrow b=c\).
Order of \(ab\) need not equal order of \(a\cdot\) order of \(b\).
Order of \(a\) = order of \(a^{-1}\) = order of \(gag^{-1}\) (conjugate).
In abelian group: order of \(ab\) divides lcm(ord \(a\), ord \(b\)).

§2 · Subgroups & Cosets

Lagrange's Theorem PYQ★★★

If \(H\) is a subgroup of finite group \(G\), then \(|H|\) divides \(|G|\).

\([G:H]=|G|/|H|\) = number of distinct left (or right) cosets of \(H\) in \(G\) = index.

Corollaries: Order of any element divides \(|G|\). If \(|G|=p\) (prime), \(G\) is cyclic. \(a^{|G|}=e\) for all \(a\in G\).

Subgroup Test

\(H\le G\) iff \(H\ne\emptyset\) and for all \(a,b\in H\): \(ab^{-1}\in H\). (One-step subgroup test.)

Intersection of subgroups is a subgroup. Union need not be.

§3 · Normal Subgroups & Quotient Groups

Normal Subgroup PYQ★★★

\(N\trianglelefteq G\) iff \(gNg^{-1}=N\) for all \(g\in G\) iff left cosets = right cosets.

Every subgroup of an abelian group is normal. Every subgroup of index 2 is normal.

Quotient Group \(G/N\)

Cosets of \(N\) in \(G\) form a group under \((aN)(bN)=abN\) iff \(N\trianglelefteq G\). Order: \(|G/N|=|G|/|N|\).

Correspondence Theorem

If \(N\trianglelefteq G\), there is a bijection between subgroups of \(G/N\) and subgroups of \(G\) containing \(N\).

§4 · Homomorphisms & Isomorphisms

Homomorphism

\(\phi:G\to H\) is a homomorphism iff \(\phi(ab)=\phi(a)\phi(b)\).

\(\ker\phi=\{g\in G:\phi(g)=e_H\}\) — always a normal subgroup of \(G\).

\(\text{Im}\,\phi\) — always a subgroup of \(H\).

First Isomorphism Theorem PYQ★★★

If \(\phi:G\to H\) is a homomorphism, then \(G/\ker\phi\cong\text{Im}\,\phi\).

Second & Third Isomorphism Theorems PYQ★★

2nd: If \(H\le G\), \(N\trianglelefteq G\): \(HN/N\cong H/(H\cap N)\).

3rd: If \(N\trianglelefteq M\trianglelefteq G\): \((G/N)/(M/N)\cong G/M\).

§5 · Cyclic & Permutation Groups

Cyclic Groups PYQ★★★

\(G=\langle a\rangle=\{a^n:n\in\mathbb{Z}\}\). Every cyclic group is abelian.

Subgroups of a cyclic group are cyclic.
Cyclic group of order \(n\) has exactly one subgroup of each order dividing \(n\).
Number of generators of \(\mathbb{Z}_n\) = \(\phi(n)\) (Euler's totient).
Every group of prime order is cyclic.
\(\mathbb{Z}_m\times\mathbb{Z}_n\) is cyclic iff \(\gcd(m,n)=1\).

Permutation Groups \(S_n\) PYQ★★★

\(|S_n|=n!\). Every finite group embeds in some \(S_n\) (Cayley's theorem).
Every permutation = product of disjoint cycles (unique up to order).
Order of permutation = lcm of lengths of disjoint cycles.
Even permutation = even number of transpositions. \(A_n\) = alternating group (even permutations), \(|A_n|=n!/2\).
\(A_n\) is simple for \(n\ge 5\) (cannot be decomposed further).

Cycle Notation Tips

\((1\,2\,3)(1\,3\,4)=\) ? Apply right to left: track where each element goes.

\((a_1\,a_2\cdots a_k)^{-1}=(a_k\,a_{k-1}\cdots a_1)\).

Conjugate: \(\sigma(a_1\,\cdots a_k)\sigma^{-1}=(\sigma(a_1)\,\cdots\,\sigma(a_k))\).

§6 · Sylow Theorems

Sylow Theorems PYQ★★★

Let \(|G|=p^a m\) with \(p\nmid m\). A Sylow \(p\)-subgroup has order \(p^a\).

Sylow I: Sylow \(p\)-subgroups exist.
Sylow II: All Sylow \(p\)-subgroups are conjugate.
Sylow III: Number of Sylow \(p\)-subgroups, \(n_p\), satisfies \(n_p\equiv 1\pmod p\) and \(n_p\mid m\).

Use: If \(n_p=1\), the Sylow \(p\)-subgroup is normal (unique ⟹ normal).

Applying Sylow to Prove Simplicity / Non-simplicity

For \(|G|=pq\) (\(p \lt q\) primes): \(G\) is not simple (the Sylow \(q\)-subgroup has \(n_q\mid p\) and \(n_q\equiv 1\pmod q\), forcing \(n_q=1\)).

For \(|G|=60=2^2\cdot3\cdot5\): \(A_5\) is the unique simple group of this order.

§7 · Rings & Ideals

Ring Axioms & Types PYQ★★

A ring \((R,+,\cdot)\): \((R,+)\) abelian group, \(\cdot\) associative, distributive over \(+\).

Commutative ring: \(ab=ba\).
Ring with unity: has multiplicative identity 1.
Division ring (skew field): every nonzero element has multiplicative inverse.
Field: commutative division ring.

Ideals PYQ★★★

\(I\subseteq R\) is an ideal if \((I,+)\le (R,+)\) and \(rI\subseteq I\) and \(Ir\subseteq I\) for all \(r\in R\).

Principal ideal: \((a)=\{ra:r\in R\}\).
Prime ideal \(P\): \(ab\in P\Rightarrow a\in P\) or \(b\in P\). \(R/P\) is integral domain.
Maximal ideal \(M\): no ideal strictly between \(M\) and \(R\). \(R/M\) is a field.
Maximal ideal ⟹ prime ideal (in commutative ring with 1).

First Isomorphism Theorem for Rings PYQ★★

If \(\phi:R\to S\) is a ring homomorphism, then \(R/\ker\phi\cong\text{Im}\,\phi\).

§8 · Integral Domains & Fields

Integral Domain & Zero Divisors

Commutative ring with 1, no zero divisors (\(ab=0\Rightarrow a=0\) or \(b=0\)).

Every field is an integral domain. Every finite integral domain is a field.

Cancellation law holds in integral domains.

Characteristic of a Ring/Field PYQ★★

Characteristic = smallest \(n\) with \(n\cdot 1=0\) (or 0 if none). In an integral domain, characteristic is 0 or prime.

Finite field \(\mathbb{F}_q\): \(q=p^n\) for prime \(p\). Exactly one field of each prime-power order (up to isomorphism).

§9 · Polynomial Rings & Factorization

Division Algorithm for Polynomials PYQ★★★

For \(f(x),g(x)\in F[x]\), \(g\ne 0\): \(\exists!\, q,r\) s.t. \(f=qg+r\) with \(\deg r < \deg g\).

Remainder theorem: Remainder when \(f(x)\) divided by \((x-a)\) is \(f(a)\).

Factor theorem: \((x-a)\mid f(x)\) iff \(f(a)=0\).

Irreducibility Tests PYQ★★★

Eisenstein's criterion: \(f(x)=a_nx^n+\cdots+a_0\in\mathbb{Z}[x]\). If prime \(p\) divides \(a_0,\ldots,a_{n-1}\) but not \(a_n\), and \(p^2\nmid a_0\), then \(f\) is irreducible over \(\mathbb{Q}\).

Reduction mod \(p\): If \(\bar f\) is irreducible mod \(p\) and \(\deg\bar f=\deg f\), then \(f\) is irreducible over \(\mathbb{Q}\).

Degree 2 or 3 polynomial: irreducible over \(F\) iff no root in \(F\).

UFD, PID, Euclidean Domain

Euclidean Domain ⟹ PID (Principal Ideal Domain) ⟹ UFD (Unique Factorization Domain).

\(\mathbb{Z}\), \(F[x]\) are Euclidean. Every PID is a UFD. \(\mathbb{Z}[x]\) is UFD but not PID.

In a UFD, irreducible = prime.