Answer: This response thoroughly evaluates five statements pertaining to group theory and field theory, as covered in the MMT-003 Algebra course. Each sub-question analyzes specific properties related to group actions, symmetric groups, field extensions, and finite fields. Detailed justifications are provided, explaining whether each statement is true or false based on fundamental algebraic principles.

Answer: The natural action of GL2(R) on M2(R) by left multiplication means that for a matrix g ∈ GL2(R) and x ∈ M2(R), the action is defined as g ⋅ x = gx. Understanding the stabilizer of an element x under this action is crucial in group theory, as it identifies all group elements that leave x unchanged. This problem explores the stabilizer for both invertible and singular matrices, highlighting fundamental properties of matrix algebra and group actions.

Answer: Let H be a finite group, P be a p-Sylow subgroup of H, and P be normal in H (P ◁ H). The order of P, denoted |P|, is the highest power of the prime p that divides the order of H, |H|. A subgroup N is normal in a group G if for every element g in G, the conjugate gNg⁻¹ is equal to N. Given that P is normal in H, a crucial consequence (as covered in topics related to Sylow's Theorems in MMT-003) is that P must be the unique p-Sylow subgroup of H. If there were any other p-Sylow subgroup P' of H, ...

Answer: As per the theory of finite abelian groups outlined in IGNOU MMT-003, we can determine the elementary divisors and invariant factors by first applying the primary decomposition theorem. We begin by expressing each component group as a direct product of cyclic groups of prime power order. First, we decompose each cyclic group: Z8 = Z(2³), Z12 = Z(2²) × Z(3), and Z15 = Z(3) × Z(5). Combining these, the given group Z8 × Z12 × Z15 becomes isomorphic to Z(2³) × Z(2²) × Z(3) × Z(3) × Z(5). Rearrang...

Answer: The quadratic polynomial x² – 11 splits into linear factors over Zp if and only if the congruence x² ≡ 11 (mod p) has solutions in Zp. This condition implies that 11 must be a quadratic residue modulo p, which is often expressed using the Legendre symbol (11/p) = 1, or that p is a prime factor of the discriminant. We must first consider special cases for the prime p. If p = 2, the polynomial becomes x² – 11 ≡ x² – 1 ≡ x² + 1 (mod 2). This equation has a solution x = 1, since 1² + 1 = 2 ≡ 0 (mod...

Answer: For a finite group G with exactly 2 conjugacy classes (k(G)=2), one class must be {e}, the identity element, of size 1 (MMT-003 Unit 2). Thus, the second class, C₂, contains all |G|-1 non-identity elements, so C₂ = G \ {e}. Elements with singleton conjugacy classes form the center Z(G). As {e} is the only such class, Z(G) must be trivial, i.e., Z(G) = {e}. For any x ≠ e, |Cl(x)| = |G| - 1. Using the class equation, |Cl(x)| = |G| / |C_G(x)|, it follows that |G| - 1 = |G| / |C_G(x)|. This implie...

Answer: No, there is no finite group with the class equation 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2. The sum of the terms in a class equation gives the order of the group, |G|. For the given equation, 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 = 14. Thus, if such a group existed, its order would be |G| = 14. From the properties of the class equation (as discussed in MMT-003, Block 2, Unit 5), the number of terms equal to '1' corresponds to the order of the center of the group, |Z(G)|. In this case, there are two '1's, impl...

Answer: The computation of `[a; b]`, which denotes the Greatest Common Divisor (GCD) of `a` and `b`, is performed using the Euclidean Algorithm. This method involves successive divisions until a zero remainder is found, with the last non-zero remainder being the GCD. **(a) [173; 211]** Using the Euclidean Algorithm: 211 = 1 ⋅ 173 + 38; 173 = 4 ⋅ 38 + 21; 38 = 1 ⋅ 21 + 17; 21 = 1 ⋅ 17 + 4; 17 = 4 ⋅ 4 + 1; 4 = 4 ⋅ 1 + 0. The last non-zero remainder is 1. Thus, GCD(173, 211) = 1. **(b) [167; 239]** Apply...

Answer: Let n = [F(a):F]. We are given that n is an odd degree, greater than 1. Our objective is to demonstrate that the field F(a²) is identical to F(a). Consider the tower of fields F ⊆ F(a²) ⊆ F(a). According to the Tower Law for field extensions, [F(a):F] = [F(a):F(a²)] * [F(a²):F]. Since 'a' is a root of the polynomial x² - a² ∈ F(a²)[x], the degree of the extension [F(a):F(a²)] can only be 1 or 2. If [F(a):F(a²)] were 2, then n = 2 * [F(a²):F], which would imply that n is an even number. This di...

Answer: The degree of a field extension, denoted [L:F], represents the dimension of L as a vector space over F. Given that α is algebraic over F of degree m, it means its minimal polynomial over F has degree m, and therefore [F(α) : F] = m. We utilize the Tower Law for field extensions, which states that for a tower of fields F ⊆ F(α) ⊆ F(α, β), the degree is multiplicative: [F(α, β) : F] = [F(α, β) : F(α)] [F(α) : F]. We already know [F(α) : F] = m. Since β is algebraic over F of degree n, its minima...

Answer: The degree of the field extension [Q(√2, ω) : Q] is determined using the Tower Law for field extensions. This law states that [Q(√2, ω) : Q] = [Q(√2, ω) : Q(√2)] * [Q(√2) : Q]. First, consider the extension [Q(√2) : Q]. The element √2 is a root of the polynomial x² - 2. This polynomial is irreducible over Q because its roots, ±√2, are not rational numbers. Therefore, the minimal polynomial for √2 over Q is x² - 2, and its degree is 2. Thus, [Q(√2) : Q] = 2. Next, consider the extension [Q(√2, ...

Answer: A polynomial ax² + bx + c over field F (where char(F) ≠ 2) is reducible if and only if it possesses roots within F. This directly implies factorability into linear terms over F. The roots are given by the quadratic formula, x = (-b ± √(b² - 4ac)) / 2a. For these roots to be elements of F, the discriminant D = b² - 4ac must be a square in F, i.e., D ∈ F*². Hence, the polynomial is irreducible if and only if it lacks roots in F. This condition is met precisely when b² – 4ac ∉ F*², establishing t...

Answer: The field F₃ consists of the elements {0, 1, 2} where arithmetic is performed modulo 3. We begin by factoring x³ – x in F₃[x]. x³ – x = x(x² – 1). Since -1 ≡ 2 (mod 3), x² – 1 = x² + 2. This factors further as (x-1)(x+1) = (x+2)(x+1) because 1 ≡ -2 and 2 ≡ -1 (mod 3). Thus, the complete factorization is x³ – x = x(x+1)(x+2). This factorization shows that x³ – x has roots 0, 1, and 2, which are all the elements of F₃. For a polynomial of degree 2 over F₃, it is irreducible if and only if it has...

Answer: Let φ: F → F be defined by φ(x) = x³ - x. We show φ is not surjective. First, φ(0) = 0 and φ(-1) = 0. If char(F) ≠ 2, then 0 ≠ -1, so φ is not injective. If char(F) = 2, then φ(0) = φ(1) = 0 and 0 ≠ 1, so φ is not injective. Since F is a finite field, a function from F to F is surjective if and only if it is injective. Therefore, φ is not surjective. This implies there exists at least one element 'a' ∈ F such that 'a' is not in the image of φ. For such an 'a', consider the polynomial P(x) = x³...

Answer: A 2n × 2n matrix M is symplectic if MᵀJM = J, where J = [0 I; -I 0] is the standard symplectic matrix. For M = [A B; C D], its transpose is Mᵀ = [Aᵀ Cᵀ; Bᵀ Dᵀ]. Block multiplication of MᵀJM equates to J, yielding the four conditions: AᵀC = CᵀA AᵀD - CᵀB = I BᵀC - DᵀA = -I BᵀD = DᵀB These derived conditions match the problem's (AC-CA=0, BD-DB=0, AD-CB=I) if A, B, C, D are symmetric matrices (i.e., Aᵀ=A, Bᵀ=B, Cᵀ=C, Dᵀ=D). Under this symmetry, AᵀC=AC (thus AC-CA=0) and BᵀD=BD (thus BD-DB=0). Add...

Answer: The following response addresses the properties of symplectic matrices in GL2(R) and their action, as requested in the sub-questions. It utilizes the definition of a symplectic matrix and properties of group actions to demonstrate the required conditions.

Answer: Let the order of the dihedral group D_n be `|D_n| = 2n`. Given `n = pʳl` where `p` is an odd prime and `p ∤ l`, the order becomes `|D_n| = 2 * pʳ * l`. By definition, a Sylow p-subgroup of D_n must have order `pʳ`, as `pʳ` is the highest power of `p` dividing `|D_n|`. The subgroup `C` is defined as `(xˡ)`. In `D_n`, `x` is an element of order `n`. The order of `xˡ` is `n / gcd(n, l)`. Substituting `n = pʳl`, we get `order(xˡ) = pʳl / gcd(pʳl, l)`. Since `p ∤ l`, `pʳ` and `l` are coprime, making...

Answer: The Dihedral group Dₙ is defined by the relations xⁿ = e, y² = e, and yx = x⁻¹y. From y² = e, we have y⁻¹ = y. We first prove the standard relation y⁻¹xʲy = x⁻ʲ, as the formula provided in the question statement appears to have an undefined variable 'k' and is generally not true for Dₙ. To prove y⁻¹xʲy = x⁻ʲ (which is yxʲy = x⁻ʲ): We use induction on j. For j=1, yx¹y = x⁻¹ (derived from yx = x⁻¹y by right-multiplying by y). Assume yxᵐy = x⁻ᵐ for some integer m ≥ 1. Then yxᵐ⁺¹y = yxᵐ(xy). Using ...

Answer: For the dihedral group Dₙ, where n is odd, its order is |Dₙ| = 2n. Since n is odd, the highest power of 2 dividing 2n is 2¹. Consequently, any Sylow 2-subgroup of Dₙ must have order 2. In Dₙ, defined by generators x (rotation of order n) and y (reflection of order 2), the elements of order 2 are precisely the reflections: xⁱy, for i = 0, 1, ..., n-1. Each such element generates a cyclic subgroup of order 2. Thus, the Sylow 2-subgroups are <y>, <xy>, <x²y>, ..., <xⁿ⁻¹y>. There are n such distinc...

Answer: To show that HN is a subgroup of Dₙ, we first establish that N is a normal subgroup of Dₙ. The subgroup N = ⟨xᵐ⟩ is a cyclic subgroup of ⟨x⟩, the group of rotations. In Dₙ, any subgroup of ⟨x⟩ that is invariant under conjugation by y is normal in Dₙ. We check yxᵐy⁻¹. Since yx = x⁻¹y, then yxᵐy⁻¹ = yxᵐy = x⁻ᵐ. As N = ⟨xᵐ⟩ has order 2ᵏ, x⁻ᵐ is equivalent to x⁽²ᵏ⁻¹⁾ᵐ, which is an element of N. Therefore, N is a normal subgroup of Dₙ. Since H = ⟨y⟩ is a subgroup and N is a normal subgroup, HN is a ...

Answer: Let n be an even integer, as implied from the previous part, specifically n = 2k where k is an odd integer (e.g., n=6, 10). The order of the dihedral group Dₙ is 2n = 4k. According to Sylow's Theorems, the order of a Sylow 2-subgroup of Dₙ is 2² = 4. These subgroups are isomorphic to the Klein 4-group (Z₂ × Z₂), as Dₙ contains no elements of order 4 from rotational elements (powers of x) for this specific 'n'. Each Sylow 2-subgroup must contain the identity 'e' and three elements of order 2. On...

Answer: The group G is defined by generators `a, b` and relations `a² = e`, `b³ = e`, and `aba⁻¹b⁻¹ = e`. These relations immediately tell us that the order of `a` is 2 and the order of `b` is 3, as `e` denotes the identity element. The relation `aba⁻¹b⁻¹ = e` can be rewritten by multiplying by `ba` on the right. This simplifies to `ab = ba`. This crucial identity shows that the generators `a` and `b` commute with each other. Since `a` and `b` commute, any element in G can be uniquely expressed as `a...

Answer: To solve the given system of congruences, we will employ the Chinese Remainder Theorem (CRT), a fundamental concept covered in modules like Unit X, Section Y of MMT-003. The CRT is applicable because the moduli (17, 19, 23) are pairwise coprime, ensuring a unique solution modulo their product. First, we simplify the congruences into the standard form x ≡ a (mod n). The first and third congruences are already in this form: x ≡ 2 (mod 17) and x ≡ 7 (mod 23). For the second congruence, 3x ≡ 4 (mod...

Answer: To demonstrate that the ring Z[√-19] is not a Unique Factorization Domain (UFD), we need to find an element that possesses two distinct factorizations into irreducible elements. Here, we analyze the integer 20 within this ring, where elements are of the form a + b√-19 for integers a, b. We first establish irreducibility using the norm function, N(a + b√-19) = a² + 19b². For an element to be reducible, it must have factors with smaller norms. For instance, to check if 2 is irreducible, we look f...