Abstract Algebra
Groups, rings, and fields with comprehensive coverage of homomorphisms, isomorphism theorems, Sylow theorems, ideals, and field extensions for GATE preparation.
Groups
Subgroups & Lagrange
Rings & Ideals
Homomorphisms
Fields
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01
Groups — Foundations
Groups are the fundamental algebraic structure encoding symmetry. We study definitions, basic properties, cyclic groups, and permutation groups.
Definition & Basic Properties
Definition — Group
A group \((G, \cdot)\) is a set \(G\) with a binary operation \(\cdot\) satisfying:
- Closure: \(a \cdot b \in G\) for all \(a,b \in G\)
- Associativity: \((a \cdot b) \cdot c = a \cdot (b \cdot c)\)
- Identity: There exists \(e \in G\) with \(e \cdot a = a \cdot e = a\) for all \(a\)
- Inverse: For each \(a \in G\), there exists \(a^{-1}\) with \(a \cdot a^{-1} = a^{-1} \cdot a = e\)
Cyclic Groups & Permutation Groups
Cyclic Group
A group \(G\) is cyclic if \(G = \langle a \rangle = \{a^n : n \in \mathbb{Z}\}\) for some \(a \in G\). Every cyclic group is abelian. Up to isomorphism, the cyclic groups are \(\mathbb{Z}\) (infinite) and \(\mathbb{Z}_n\) (finite, order \(n\)).
Theorem — Subgroups of Cyclic Groups
Every subgroup of a cyclic group is cyclic. If \(G = \langle a \rangle\) has order \(n\), then for each divisor \(d\) of \(n\), there is a unique subgroup of order \(d\), namely \(\langle a^{n/d} \rangle\).
Symmetric Group \(S_n\)
The symmetric group \(S_n\) is the group of all permutations of \(\{1,2,\ldots,n\}\) under composition. \(|S_n| = n!\). Every permutation can be written as a product of disjoint cycles and also as a product of transpositions.
★ Example
Find the order of the permutation \(\sigma = (1\;2\;3)(4\;5)\) in \(S_5\).
The cycle \((1\;2\;3)\) has order 3 and \((4\;5)\) has order 2. Since they are disjoint:
\[\text{ord}(\sigma) = \text{lcm}(3,2) = 6\]
02
Subgroups, Lagrange’s Theorem & Sylow Theorems
Lagrange's theorem constrains subgroup orders; normal subgroups enable quotient groups; Sylow theorems characterize p-subgroups of finite groups.
Lagrange’s Theorem & Cosets
Lagrange’s Theorem
If \(H\) is a subgroup of a finite group \(G\), then \(|H|\) divides \(|G|\). The index is \([G:H] = |G|/|H|\).
Corollaries: The order of every element divides \(|G|\). Every group of prime order is cyclic. \(a^{|G|} = e\) for all \(a \in G\).
Normal Subgroups & Quotient Groups
Normal Subgroup
A subgroup \(N \le G\) is normal (written \(N \trianglelefteq G\)) if \(gNg^{-1} = N\) for all \(g \in G\). Equivalently, \(gN = Ng\) for all \(g\).
Quotient Group
If \(N \trianglelefteq G\), the set of cosets \(G/N = \{gN : g \in G\}\) forms a group under \((aN)(bN) = (ab)N\), with \(|G/N| = [G:N]\).
Sylow Theorems
Sylow Theorems
Let \(|G| = p^a m\) where \(p \nmid m\). Then:
- Sylow I: \(G\) has a subgroup of order \(p^a\) (a Sylow \(p\)-subgroup).
- Sylow II: All Sylow \(p\)-subgroups are conjugate.
- Sylow III: The number \(n_p\) of Sylow \(p\)-subgroups satisfies \(n_p \equiv 1 \pmod{p}\) and \(n_p \mid m\).
★ Example
Show that a group of order 15 is cyclic.
\(|G| = 15 = 3 \cdot 5\). By Sylow III: \(n_3 \mid 5\) and \(n_3 \equiv 1\pmod{3}\), so \(n_3 = 1\). Similarly \(n_5 \mid 3\) and \(n_5 \equiv 1\pmod{5}\), so \(n_5 = 1\).
Both Sylow subgroups are normal. Since \(\gcd(3,5)=1\), \(G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}\). Thus \(G\) is cyclic.
Both Sylow subgroups are normal. Since \(\gcd(3,5)=1\), \(G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}\). Thus \(G\) is cyclic.
03
Homomorphisms & Isomorphism Theorems
Group homomorphisms preserve structure. The isomorphism theorems are the fundamental tools for relating groups, their subgroups, and quotients.
Group Homomorphisms
Definition — Homomorphism
A map \(\phi: G \to H\) is a group homomorphism if \(\phi(ab) = \phi(a)\phi(b)\) for all \(a,b \in G\). The kernel is \(\ker\phi = \{g \in G : \phi(g) = e_H\}\) and the image is \(\text{Im}(\phi) = \phi(G)\).
Key fact: \(\ker\phi \trianglelefteq G\) and \(\text{Im}(\phi) \le H\). A homomorphism is injective iff \(\ker\phi = \{e\}\).
Isomorphism Theorems
First Isomorphism Theorem
If \(\phi: G \to H\) is a homomorphism, then:
\[G/\ker\phi \cong \text{Im}(\phi)\]
Second Isomorphism Theorem
If \(H \le G\) and \(N \trianglelefteq G\), then \(HN/N \cong H/(H \cap N)\).
Third Isomorphism Theorem
If \(N \trianglelefteq G\) and \(N \le M \trianglelefteq G\), then \((G/N)/(M/N) \cong G/M\).
★ Example
Consider \(\phi: \mathbb{Z} \to \mathbb{Z}_n\) defined by \(\phi(k) = k \bmod n\). Apply the First Isomorphism Theorem.
\(\phi\) is a surjective homomorphism with \(\ker\phi = n\mathbb{Z}\). By the First Isomorphism Theorem:
\[\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n\]
This confirms the standard construction of \(\mathbb{Z}_n\) as a quotient group.
04
Rings, Ideals & Polynomial Rings
Rings generalize integer arithmetic. Ideals play the role of normal subgroups, enabling quotient ring constructions.
Rings & Ideals
Definition — Ring
A ring \((R, +, \cdot)\) is an abelian group under \(+\) with an associative multiplication that distributes over addition. If \(\cdot\) is commutative and there exists \(1 \ne 0\), we call \(R\) a commutative ring with unity.
Ideal
A subset \(I \subseteq R\) is an ideal if \((I,+) \le (R,+)\) and \(rI \subseteq I\), \(Ir \subseteq I\) for all \(r \in R\). The quotient ring \(R/I\) has elements \(r + I\) with operations \((a+I)+(b+I)=(a+b)+I\) and \((a+I)(b+I)=ab+I\).
Prime & Maximal Ideals
- An ideal \(P\) is prime if \(ab \in P \Rightarrow a \in P\) or \(b \in P\). Equivalently, \(R/P\) is an integral domain.
- An ideal \(M\) is maximal if there is no ideal strictly between \(M\) and \(R\). Equivalently, \(R/M\) is a field.
PID, UFD, ED
Definitions
- Principal Ideal Domain (PID): An integral domain where every ideal is principal, i.e., \(I = (a)\) for some \(a\).
- Unique Factorization Domain (UFD): Every non-zero non-unit factors uniquely (up to order and associates) into irreducibles.
- Euclidean Domain (ED): An integral domain with a Euclidean function allowing division with remainder.
★ Example
Show that \(\mathbb{Z}[x]\) is a UFD but not a PID.
\(\mathbb{Z}\) is a UFD, and by Gauss's Lemma, \(\mathbb{Z}[x]\) is also a UFD.
However, the ideal \((2, x) = \{2f(x) + xg(x) : f,g \in \mathbb{Z}[x]\}\) is not principal. If \((2,x) = (d)\), then \(d \mid 2\) and \(d \mid x\), forcing \(d\) to be a unit, but \((2,x) \ne \mathbb{Z}[x]\) since \(1 \notin (2,x)\). Contradiction. So \(\mathbb{Z}[x]\) is not a PID.
However, the ideal \((2, x) = \{2f(x) + xg(x) : f,g \in \mathbb{Z}[x]\}\) is not principal. If \((2,x) = (d)\), then \(d \mid 2\) and \(d \mid x\), forcing \(d\) to be a unit, but \((2,x) \ne \mathbb{Z}[x]\) since \(1 \notin (2,x)\). Contradiction. So \(\mathbb{Z}[x]\) is not a PID.
05
Fields & Field Extensions
Fields are rings where every nonzero element is invertible. Field extensions underlie Galois theory and the study of finite fields.
Field Extensions
Definition — Field Extension
If \(F \subseteq K\) are fields, then \(K\) is a field extension of \(F\), written \(K/F\). The degree \([K:F] = \dim_F K\) is the dimension of \(K\) as a vector space over \(F\).
Tower Law
If \(F \subseteq K \subseteq L\), then \([L:F] = [L:K][K:F]\).
★ Example
Find \([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\).
\([\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\) since \(x^2-2\) is irreducible over \(\mathbb{Q}\).
\([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})] = 2\) since \(x^2-3\) is irreducible over \(\mathbb{Q}(\sqrt{2})\) (as \(\sqrt{3} \notin \mathbb{Q}(\sqrt{2})\)).
By the Tower Law: \([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = 2 \times 2 = 4\).
\([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})] = 2\) since \(x^2-3\) is irreducible over \(\mathbb{Q}(\sqrt{2})\) (as \(\sqrt{3} \notin \mathbb{Q}(\sqrt{2})\)).
By the Tower Law: \([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = 2 \times 2 = 4\).
Finite Fields & Splitting Fields
Classification of Finite Fields
For every prime \(p\) and positive integer \(n\), there exists a unique (up to isomorphism) finite field \(\mathbb{F}_{p^n}\) of order \(p^n\). It is the splitting field of \(x^{p^n} - x\) over \(\mathbb{F}_p\). The multiplicative group \(\mathbb{F}_{p^n}^*\) is cyclic of order \(p^n - 1\).
Splitting Field
The splitting field of a polynomial \(f(x) \in F[x]\) over \(F\) is the smallest extension of \(F\) in which \(f\) factors completely into linear factors. It is unique up to isomorphism.
★ Example
Find the splitting field of \(x^4 - 1\) over \(\mathbb{F}_3\).
\(x^4 - 1 = (x-1)(x+1)(x^2+1)\) over \(\mathbb{F}_3\). Check: \(x^2+1\) has no roots in \(\mathbb{F}_3\) (since \(0^2+1=1\), \(1^2+1=2\), \(2^2+1=2\), none zero). So \(x^2+1\) is irreducible over \(\mathbb{F}_3\).
The splitting field is \(\mathbb{F}_3[x]/(x^2+1) \cong \mathbb{F}_9\), of degree 2 over \(\mathbb{F}_3\).
The splitting field is \(\mathbb{F}_3[x]/(x^2+1) \cong \mathbb{F}_9\), of degree 2 over \(\mathbb{F}_3\).
★ Key Takeaways
- Lagrange's theorem: \(|H|\) divides \(|G|\); every group of prime order is cyclic.
- Sylow theorems determine the number and structure of p-subgroups; key for proving groups are not simple.
- First Isomorphism Theorem: \(G/\ker\phi \cong \text{Im}(\phi)\) is the most used structural result.
- ED \(\Rightarrow\) PID \(\Rightarrow\) UFD; \(\mathbb{Z}[x]\) is UFD but not PID.
- Finite fields \(\mathbb{F}_{p^n}\) exist uniquely for each prime power; their multiplicative group is cyclic.
📝 Practice Problems
Problem 1
Find all subgroups of \(\mathbb{Z}_{12}\).
Show Solution ▼
Divisors of 12: 1, 2, 3, 4, 6, 12. The subgroups are \(\langle 0 \rangle = \{0\}\), \(\langle 6 \rangle = \{0,6\}\), \(\langle 4 \rangle = \{0,4,8\}\), \(\langle 3 \rangle = \{0,3,6,9\}\), \(\langle 2 \rangle = \{0,2,4,6,8,10\}\), and \(\mathbb{Z}_{12}\). One subgroup for each divisor.
Problem 2
Let \(G\) be a group of order 35. Prove that \(G\) is cyclic.
Show Solution ▼
\(|G|=35=5 \times 7\). Sylow: \(n_5 \mid 7\) and \(n_5 \equiv 1\pmod{5}\), so \(n_5 = 1\). \(n_7 \mid 5\) and \(n_7 \equiv 1\pmod{7}\), so \(n_7 = 1\). Both Sylow subgroups are normal, with coprime orders 5 and 7. So \(G \cong \mathbb{Z}_5 \times \mathbb{Z}_7 \cong \mathbb{Z}_{35}\).
Problem 3
Is the ideal \((x)\) in \(\mathbb{Z}[x]\) prime? Maximal?
Show Solution ▼
\(\mathbb{Z}[x]/(x) \cong \mathbb{Z}\). Since \(\mathbb{Z}\) is an integral domain but not a field, \((x)\) is prime but not maximal.
Problem 4
Find the order of the element \(\bar{3}\) in \((\mathbb{Z}/7\mathbb{Z})^*\).
Show Solution ▼
\(|(\mathbb{Z}/7\mathbb{Z})^*| = 6\). Compute: \(3^1=3\), \(3^2=2\), \(3^3=6\), \(3^4=4\), \(3^5=5\), \(3^6=1\). The order is 6, so \(\bar{3}\) is a generator of the cyclic group \((\mathbb{Z}/7\mathbb{Z})^*\).
Problem 5
Find the degree \([\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]\) and determine if this is a splitting field of \(x^3-2\).
Show Solution ▼
\(x^3-2\) is irreducible over \(\mathbb{Q}\) (Eisenstein with \(p=2\)), so \([\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3\). However, \(x^3-2\) has roots \(\sqrt[3]{2}\), \(\sqrt[3]{2}\omega\), \(\sqrt[3]{2}\omega^2\) where \(\omega = e^{2\pi i/3}\). Since \(\omega \notin \mathbb{R}\) and \(\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}\), this is not the splitting field. The splitting field is \(\mathbb{Q}(\sqrt[3]{2}, \omega)\) with degree 6 over \(\mathbb{Q}\).
🎯 Interactive Quiz
1. The number of elements in the symmetric group \(S_4\) is:
A 4
B 12
C 24
D 16
2. If \(G\) is a group of order 7, then \(G\) is:
A Non-abelian
B Cyclic
C Not simple
D Isomorphic to \(S_7\)
3. In a commutative ring with unity, a maximal ideal \(M\) has the property that \(R/M\) is:
A An integral domain
B A field
C A ring with zero divisors
D The trivial ring
4. The number of Sylow 3-subgroups of a group of order 12 can be:
A 2
B 3
C 1 or 4
D 6
5. The multiplicative group of the finite field \(\mathbb{F}_{16}\) is:
A Cyclic of order 15
B Cyclic of order 16
C \(\mathbb{Z}_3 \times \mathbb{Z}_5\) (non-cyclic)
D Not cyclic