Abstract Algebra
Group theory, ring theory, field extensions, Galois theory, and module basics at the CSIR NET level with emphasis on Sylow theorems, classification results, and Galois correspondences.
0% complete6 units
01
Group Theory Fundamentals
Groups, subgroups, normal subgroups, homomorphisms, isomorphism theorems, group actions, and semidirect products.
Isomorphism Theorems & Group Actions
First Isomorphism Theorem
If $\phi: G \to H$ is a group homomorphism, then $G / \ker \phi \cong \text{im}\, \phi$. In particular, $|G| = |\ker \phi| \cdot |\text{im}\, \phi|$ for finite groups.
Second Isomorphism Theorem
If $H \leq G$ and $N \trianglelefteq G$, then $HN/N \cong H/(H \cap N)$.
Group Action
A group $G$ acts on a set $X$ via $\alpha: G \times X \to X$ satisfying $e \cdot x = x$ and $(gh) \cdot x = g \cdot (h \cdot x)$. The orbit of $x$ is $Gx = \{g \cdot x : g \in G\}$ and the stabilizer is $G_x = \{g \in G : g \cdot x = x\}$.
Orbit-Stabilizer Theorem
$|G| = |Gx| \cdot |G_x|$ for any $x \in X$.
Class Equation
For a finite group $G$ acting on itself by conjugation: $|G| = |Z(G)| + \sum [G : C_G(x_i)]$, where the sum is over representatives of conjugacy classes of size $> 1$.
Semidirect Products
Definition
Given $N \trianglelefteq G$ and $H \leq G$ with $G = NH$ and $N \cap H = \{e\}$, we write $G = N \rtimes H$. This is determined by a homomorphism $\phi: H \to \text{Aut}(N)$, with multiplication $(n_1, h_1)(n_2, h_2) = (n_1 \cdot \phi(h_1)(n_2), h_1 h_2)$.
📝 Example
Show that $S_3 \cong \mathbb{Z}_3 \rtimes \mathbb{Z}_2$.
Take $N = \langle (123) \rangle \cong \mathbb{Z}_3$ (normal in $S_3$) and $H = \langle (12) \rangle \cong \mathbb{Z}_2$. Then $S_3 = NH$, $N \cap H = \{e\}$, and $H$ acts on $N$ by $\phi((12))((123)) = (132) = (123)^{-1}$, i.e., inversion. So $S_3 \cong \mathbb{Z}_3 \rtimes_\phi \mathbb{Z}_2$.
02
Sylow Theorems & Classification
Sylow theorems, applications to classifying groups of small order, and the fundamental theorem of finitely generated abelian groups.
Sylow Theorems
Sylow Theorems
Let $|G| = p^a m$ with $\gcd(p, m) = 1$. 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
Prove that every group of order $15$ is cyclic.
$|G| = 15 = 3 \cdot 5$. By Sylow III: $n_3 | 5$ and $n_3 \equiv 1 \pmod{3}$, so $n_3 = 1$. Similarly $n_5 | 3$ and $n_5 \equiv 1 \pmod{5}$, so $n_5 = 1$. Both Sylow subgroups are normal. $G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}$ since $\gcd(3,5) = 1$.
Finitely Generated Abelian Groups
Fundamental Theorem
Every finitely generated abelian group is isomorphic to $\mathbb{Z}^r \times \mathbb{Z}_{d_1} \times \cdots \times \mathbb{Z}_{d_k}$ where $r \geq 0$ and $d_1 | d_2 | \cdots | d_k$ (invariant factor decomposition). Equivalently, by prime factorization: $\mathbb{Z}^r \times \mathbb{Z}_{p_1^{a_1}} \times \cdots \times \mathbb{Z}_{p_s^{a_s}}$ (elementary divisor decomposition).
📝 Example
List all abelian groups of order $72 = 2^3 \cdot 3^2$ up to isomorphism.
Partitions of $3$: $\{3\}, \{2,1\}, \{1,1,1\}$. Partitions of $2$: $\{2\}, \{1,1\}$. So the groups are: $\mathbb{Z}_8 \times \mathbb{Z}_9$, $\mathbb{Z}_8 \times \mathbb{Z}_3^2$, $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9$, $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3^2$, $\mathbb{Z}_2^3 \times \mathbb{Z}_9$, $\mathbb{Z}_2^3 \times \mathbb{Z}_3^2$. That is $3 \times 2 = 6$ groups total.
03
Ring Theory
Ideals, quotient rings, PID, UFD, Euclidean domains, polynomial rings, and localization.
Ideals & Quotient Rings
Definition
An ideal $I$ of a ring $R$ is prime if $ab \in I$ implies $a \in I$ or $b \in I$. It is maximal if $I \neq R$ and no ideal lies strictly between $I$ and $R$.
Characterization
In a commutative ring with $1$: $I$ is prime iff $R/I$ is an integral domain. $I$ is maximal iff $R/I$ is a field. Consequently, every maximal ideal is prime.
PID, UFD, and Euclidean Domains
Domain Hierarchy
Euclidean Domain $\implies$ PID $\implies$ UFD $\implies$ Integral Domain. The converse implications are false in general.
Key examples: $\mathbb{Z}$ and $F[x]$ (for a field $F$) are Euclidean domains. $\mathbb{Z}[x]$ is a UFD but not a PID (the ideal $(2, x)$ is not principal). $\mathbb{Z}[\sqrt{-5}]$ is not a UFD: $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$.
Gauss's Lemma & Eisenstein's Criterion
Gauss: If $R$ is a UFD and $f \in R[x]$ is primitive, then $f$ is irreducible in $R[x]$ iff it is irreducible in $\text{Frac}(R)[x]$.
Eisenstein: If $f(x) = a_n x^n + \cdots + a_0 \in \mathbb{Z}[x]$ and there is a prime $p$ with $p \nmid a_n$, $p | a_i$ for $i < n$, and $p^2 \nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.
Eisenstein: If $f(x) = a_n x^n + \cdots + a_0 \in \mathbb{Z}[x]$ and there is a prime $p$ with $p \nmid a_n$, $p | a_i$ for $i < n$, and $p^2 \nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.
📝 Example
Prove that $x^4 + 1$ is irreducible over $\mathbb{Q}$.
Substitute $x \mapsto x + 1$: $(x+1)^4 + 1 = x^4 + 4x^3 + 6x^2 + 4x + 2$. Apply Eisenstein with $p = 2$: $2 \nmid 1$ (leading coeff), $2 | 4, 6, 4$, and $4 \nmid 2$. By Eisenstein, the shifted polynomial is irreducible over $\mathbb{Q}$, hence so is $x^4 + 1$.
04
Field Extensions
Algebraic and transcendental extensions, degree of extensions, splitting fields, and finite fields.
Algebraic Extensions
Definition
An element $\alpha$ is algebraic over $F$ if it is a root of some nonzero polynomial in $F[x]$. The minimal polynomial $m_\alpha(x)$ is the unique monic irreducible polynomial in $F[x]$ with $m_\alpha(\alpha) = 0$. The degree $[F(\alpha):F] = \deg m_\alpha$.
Tower Law
If $F \subseteq K \subseteq L$ are fields, then $[L:F] = [L:K] \cdot [K:F]$. Consequently, if $[L:F]$ is prime, there are no intermediate fields.
Finite Fields
For every prime power $q = p^n$, there exists a unique (up to isomorphism) field $\mathbb{F}_q$ with $q$ elements. $\mathbb{F}_q^*$ is cyclic of order $q - 1$, and $\mathbb{F}_q$ is the splitting field of $x^q - x$ over $\mathbb{F}_p$.
📝 Example
Find $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]$.
$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$ (minimal poly $x^2-2$). Is $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$? If $\sqrt{3} = a + b\sqrt{2}$, then $3 = a^2 + 2b^2 + 2ab\sqrt{2}$, forcing $ab = 0$. If $b = 0$: $\sqrt{3} = a \in \mathbb{Q}$, impossible. If $a = 0$: $3 = 2b^2$, giving $b^2 = 3/2 \notin \mathbb{Q}$, impossible. So $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})] = 2$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = 4$.
05
Galois Theory
Galois extensions, the fundamental theorem, computing Galois groups, solvability by radicals, and applications.
The Fundamental Theorem of Galois Theory
Definition
An extension $L/F$ is Galois if it is normal (splitting field of a polynomial) and separable (minimal polynomials have distinct roots). The Galois group $\text{Gal}(L/F)$ is the group of $F$-automorphisms of $L$.
Fundamental Theorem of Galois Theory
If $L/F$ is a finite Galois extension with $G = \text{Gal}(L/F)$, there is an inclusion-reversing bijection between subgroups $H \leq G$ and intermediate fields $F \subseteq K \subseteq L$, given by $H \mapsto L^H$ (fixed field) and $K \mapsto \text{Gal}(L/K)$. Moreover, $[L:K] = |H|$ and $K/F$ is Galois iff $H \trianglelefteq G$, in which case $\text{Gal}(K/F) \cong G/H$.
📝 Example
Find $\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$.
$L = \mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $(x^2-2)(x^2-3)$. $[L:\mathbb{Q}] = 4$, so $|\text{Gal}| = 4$. The automorphisms are determined by $\sigma(\sqrt{2}) = \pm\sqrt{2}$, $\sigma(\sqrt{3}) = \pm\sqrt{3}$, giving $4$ elements: $\text{id}, \sigma_1: \sqrt{2} \mapsto -\sqrt{2}, \sigma_2: \sqrt{3} \mapsto -\sqrt{3}, \sigma_3: \sqrt{2} \mapsto -\sqrt{2}, \sqrt{3} \mapsto -\sqrt{3}$. Each has order $2$, so $\text{Gal} \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (Klein four-group). The intermediate fields correspond to the three subgroups of order $2$: $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(\sqrt{6})$.
Solvability by Radicals
Galois's Theorem
A polynomial $f \in F[x]$ is solvable by radicals if and only if its Galois group $\text{Gal}(L/F)$ (where $L$ is the splitting field) is a solvable group. Since $S_n$ is not solvable for $n \geq 5$, the general polynomial of degree $\geq 5$ is not solvable by radicals.
06
Module Theory Basics
Modules over rings, free modules, and the structure theorem for finitely generated modules over a PID.
Modules & Structure Theorem
Definition
A module over a ring $R$ is an abelian group $M$ with a scalar multiplication $R \times M \to M$ satisfying the usual axioms. Modules generalize vector spaces (where $R$ is a field) and abelian groups (where $R = \mathbb{Z}$).
Structure Theorem (Modules over PID)
Every finitely generated module over a PID $R$ is isomorphic to $R^r \oplus R/(d_1) \oplus \cdots \oplus R/(d_k)$ where $d_1 | d_2 | \cdots | d_k$ are nonzero non-units (invariant factors). This unifies the fundamental theorem of abelian groups ($R = \mathbb{Z}$) and Jordan/rational canonical forms ($R = F[x]$).
When $R = F[x]$ and $M = V$ is a finite-dimensional $F$-vector space viewed as an $F[x]$-module via $x \cdot v = Tv$ for a linear operator $T$, the structure theorem yields the rational canonical form of $T$.
★ Key Takeaways
- Sylow theorems are the primary tool for analyzing groups of given order: count Sylow subgroups using divisibility constraints
- The chain ED $\Rightarrow$ PID $\Rightarrow$ UFD is strict; know the counterexamples ($\mathbb{Z}[x]$, $\mathbb{Z}[\sqrt{-5}]$)
- Eisenstein's criterion (with shifts) is the go-to tool for proving irreducibility over $\mathbb{Q}$
- The Fundamental Theorem of Galois Theory converts field extension problems into group theory problems
- The insolvability of the quintic follows from $S_5$ being non-solvable
- The structure theorem for modules over a PID unifies abelian group classification and matrix canonical forms
✍ Practice Problems
Problem 1
Prove that no group of order $56 = 2^3 \cdot 7$ is simple.
Show Solution ▼
$n_7 | 8$ and $n_7 \equiv 1 \pmod{7}$, so $n_7 \in \{1, 8\}$. If $n_7 = 1$, the unique Sylow $7$-subgroup is normal and we are done. If $n_7 = 8$, these give $8 \times 6 = 48$ elements of order $7$. The remaining $56 - 48 = 8$ elements must form the unique Sylow $2$-subgroup (since we need exactly $8$ elements and $|P_2| = 8$), so $n_2 = 1$ and this Sylow $2$-subgroup is normal. Either way, $G$ is not simple.
Problem 2
Show that $\mathbb{Z}[i]$ (Gaussian integers) is a Euclidean domain with norm $N(a + bi) = a^2 + b^2$.
Show Solution ▼
Given $\alpha, \beta \in \mathbb{Z}[i]$ with $\beta \neq 0$, write $\alpha/\beta = r + si \in \mathbb{Q}(i)$. Choose $m, n \in \mathbb{Z}$ with $|r - m| \leq 1/2$ and $|s - n| \leq 1/2$. Let $q = m + ni$ and $\rho = \alpha - q\beta$. Then $N(\rho) = N(\alpha - q\beta) = N(\beta) \cdot N(\alpha/\beta - q) = N(\beta)((r-m)^2 + (s-n)^2) \leq N(\beta)(1/4 + 1/4) = N(\beta)/2 < N(\beta)$.
Problem 3
Find the Galois group of $x^4 - 2$ over $\mathbb{Q}$.
Show Solution ▼
The roots are $\sqrt[4]{2}, -\sqrt[4]{2}, i\sqrt[4]{2}, -i\sqrt[4]{2}$, i.e., $\sqrt[4]{2} \cdot \zeta$ where $\zeta$ ranges over $\{1, -1, i, -i\}$. The splitting field is $\mathbb{Q}(\sqrt[4]{2}, i)$ with $[\mathbb{Q}(\sqrt[4]{2}, i):\mathbb{Q}] = [\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] \cdot [\mathbb{Q}(\sqrt[4]{2}, i):\mathbb{Q}(\sqrt[4]{2})] = 4 \cdot 2 = 8$. The Galois group has order $8$ and is isomorphic to the dihedral group $D_4$, generated by $\sigma: \sqrt[4]{2} \mapsto i\sqrt[4]{2}, i \mapsto i$ and $\tau: \sqrt[4]{2} \mapsto \sqrt[4]{2}, i \mapsto -i$.
Problem 4
Determine all maximal ideals of $\mathbb{Z}[x]/(x^2 + 1)$.
Show Solution ▼
$\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$ (Gaussian integers). Maximal ideals of $\mathbb{Z}[i]$ correspond to primes: (1) $(1+i)$ (above $p=2$, since $2 = -i(1+i)^2$), (2) $(\pi)$ where $\pi$ is a Gaussian prime above a rational prime $p \equiv 1 \pmod{4}$ (such primes split: $p = \pi\bar{\pi}$), (3) $(p)$ for rational primes $p \equiv 3 \pmod{4}$ (these remain prime in $\mathbb{Z}[i]$).
Problem 5
How many elements of order $7$ are there in $\mathbb{Z}_{21} \times \mathbb{Z}_{35}$?
Show Solution ▼
$\mathbb{Z}_{21} \times \mathbb{Z}_{35} \cong \mathbb{Z}_3 \times \mathbb{Z}_7 \times \mathbb{Z}_5 \times \mathbb{Z}_7$. Elements of order $7$ have trivial $\mathbb{Z}_3$ and $\mathbb{Z}_5$ components and at least one nontrivial $\mathbb{Z}_7$ component. Elements of the $\mathbb{Z}_7 \times \mathbb{Z}_7$ subgroup number $49$; those with order dividing $1$ number $1$. So elements of order exactly $7$: $49 - 1 = 48$.
🎯 Test Your Understanding
1. The number of groups of order $4$ (up to isomorphism) is:
A 1
B 2
C 3
D 4
2. In the ring $\mathbb{Z}[x]$, the ideal $(2, x)$ is:
A Principal
B Maximal
C Equal to $\mathbb{Z}[x]$
D Not prime
3. The splitting field of $x^3 - 2$ over $\mathbb{Q}$ has degree:
A 3
B 6
C 9
D 12
4. The number of elements in the finite field $\mathbb{F}_{27}^*$ (multiplicative group) is:
A 27
B 26
C 24
D 9
5. Which of the following rings is a PID?
A $\mathbb{Z}[x]$
B $\mathbb{Q}[x]$
C $\mathbb{Z}[x, y]$
D $\mathbb{Z}[\sqrt{-5}]$