Check whether or not Z[x]/ is a field.

To determine if Z[x]/ is a field, we refer to a fundamental theorem from Unit 3/4 of BMTC-134: A quotient ring R/I is a field if and only if R is a commutative ring with unity and I is a maximal ideal in R. Here, R = Z[x] and I = . Let f(x) = 8x³ + 6x² - 9x + 24. For Z[x]/ to be a field, the ideal must be maximal in Z[x]. However, a principal ideal generated by a non-constant polynomial in Z[x] is never maximal. This is because Z[x] is not a Princi...

BMTC-134 Solved Assignment January 2026 | ALGEBRA | IGNOU BSCG

Q: Which of the following statements are true? Give reasons for your answers.

Answer available — download the full PDF for complete details.

Q: Let G be an infinite group such that for any non-trivial subgroup H of G, |G : H| < ". Then prove that

Answer available — download the full PDF for complete details.

Q: Prove that a cyclic group with only one generator can have at most 2 elements.

Let G be a cyclic group generated by a single element, 'a'. This means G = ⟨a⟩. A key property of cyclic groups, as discussed in group theory (e.g., BMTC-134, Block 2, Unit 5 on Cyclic Groups), is that if 'a' is a generator, its inverse, a⁻¹, is also a generator. Since G possesses *only one* generator, it logically follows that this unique generator 'a' must be identical to its inverse. That is, a = a⁻¹. Multiplying both sides of this equation by 'a' yields a² = a * a⁻¹, which simplifies to a²

Q: Using Cayley's theorem, find the permutation group to which a cyclic group of order 12 is isomorphic.

Cayley's Theorem, a fundamental result in group theory discussed in Unit 2 of BMTC-134, states that every finite group is isomorphic to a permutation group. More precisely, if G is a group of finite order n, then G is isomorphic to a subgroup of the symmetric group S_n. This theorem allows us to represent any abstract group concretely as a group of permutations, which are bijective functions from a set to itself. In this specific problem, we are given a cyclic group of order 12. A cyclic group

Q: Let τ be a fixed odd permutation in S10. Show that every odd permutation in S10 is a product of τ and some permutation in A10.

Let τ be a fixed odd permutation in S₁₀. We need to show that any arbitrary odd permutation σ in S₁₀ can be written as a product of τ and some permutation π in A₁₀, i.e., σ = τπ. Since τ is an odd permutation, its inverse τ⁻¹ is also an odd permutation. This is a fundamental property, as the sign of a permutation and its inverse are the same, i.e., sgn(τ⁻¹) = sgn(τ). Consider the permutation π = τ⁻¹σ. We determine the parity of π. We know that sgn(π) = sgn(τ⁻¹σ) = sgn(τ⁻¹) × sgn(σ). Since both

Q1. Which of the following statements are true? Give reasons for your answers.

i)) If a group G is isomorphic to one of its proper subgroups, then G = Z. (100 words)
ii)) If x and y are elements of a non-abelian group (G, *) such that x * y = y * x, then x =e or y = e, where e is the identity of G with respect to *. (100 words)
iii)) There exists a unique non-abelian group of prime order. (100 words)
iv)) If (a, b) " A × A, where A is a group, then o((a, b)) = o(a)o(b). (100 words)
v)) If H and K are normal subgroups of a group G, then hk = khvh " H, k " Κ. (100 words)

Key Points:

Not all groups isomorphic to their proper subgroups are Z; counterexamples like dyadic rationals exist.
Non-abelian groups can have non-identity elements that commute; e.g., i and -1 in Q₈.
Groups of prime order are always cyclic and thus abelian; no non-abelian group of prime order exists.
Order of (a,b) in A×B is lcm(o(a), o(b)), not o(a)o(b), unless o(a) and o(b) are coprime.

Answer:

Q2a. Prove that every non-trivial subgroup of a cyclic group has finite index. Hence prove that (Q, +) is not cyclic.

Key Points:

Every subgroup of a cyclic group G = ⟨a⟩ is cyclic, generated by an element aᵏ.
For a non-trivial subgroup H = ⟨aᵏ⟩, the exponent k is a non-zero positive integer.
The index [G:H] is the number of distinct cosets, specifically a⁰H, a¹H, ..., aᵏ⁻¹H.
The index [G:H] equals k, which is a finite integer, proving the first part.

Answer: To prove that every non-trivial subgroup of a cyclic group has a finite index, let G be an arbitrary cyclic group generated by an element 'a', so G = ⟨a⟩. Let H be a non-trivial subgroup of G. A fundamental property of cyclic groups, often discussed in courses like BMTC-134, states that every subgroup of a cyclic group is also cyclic. Therefore, H must be cyclic and generated by some element of G. Since H is a subgroup of G, H = ⟨aᵏ⟩ for some integer k. As H is non-trivial (meaning H ≠ {e}, whe...

Q2b. Let G be an infinite group such that for any non-trivial subgroup H of G, |G : H| < ". Then prove that

i)) H < G (125 words)
ii)) If g " G, g " e, then o(g) is infinite. (125 words)

Key Points:

An infinite group G has an infinite number of elements.
The product formula |G| = |G : H| × |H| relates group order, subgroup index, and subgroup order.
If an infinite group G has a subgroup H of finite index, then H must be an infinite subgroup.
A non-trivial cyclic subgroup ⟨g⟩ generated by g ≠ e has finite index in G.

Answer:

Q2c. Prove that a cyclic group with only one generator can have at most 2 elements.

Key Points:

In a cyclic group G=⟨a⟩, if 'a' is a generator, then a⁻¹ is also a generator.
If G has only one generator, 'a' must be identical to its inverse (a = a⁻¹).
The condition a = a⁻¹ simplifies to a² = e (where 'e' is the identity element).
This implies the order of the generator o(a) is either 1 or 2.

Answer: Let G be a cyclic group generated by a single element, 'a'. This means G = ⟨a⟩. A key property of cyclic groups, as discussed in group theory (e.g., BMTC-134, Block 2, Unit 5 on Cyclic Groups), is that if 'a' is a generator, its inverse, a⁻¹, is also a generator. Since G possesses *only one* generator, it logically follows that this unique generator 'a' must be identical to its inverse. That is, a = a⁻¹. Multiplying both sides of this equation by 'a' yields a² = a * a⁻¹, which simplifies to a² ...

Q3a. Using Cayley's theorem, find the permutation group to which a cyclic group of order 12 is isomorphic.

Key Points:

Cayley's Theorem states every group is isomorphic to a permutation group.
A group G of finite order 'n' is isomorphic to a subgroup of S_n.
The given cyclic group has an order of 12 (n=12).
Applying Cayley's Theorem, the group is isomorphic to a subgroup of S_12.

Answer: Cayley's Theorem, a fundamental result in group theory discussed in Unit 2 of BMTC-134, states that every finite group is isomorphic to a permutation group. More precisely, if G is a group of finite order n, then G is isomorphic to a subgroup of the symmetric group S_n. This theorem allows us to represent any abstract group concretely as a group of permutations, which are bijective functions from a set to itself. In this specific problem, we are given a cyclic group of order 12. A cyclic group ...

Q3b. Let τ be a fixed odd permutation in S10. Show that every odd permutation in S10 is a product of τ and some permutation in A10.

Key Points:

An odd permutation's inverse is also an odd permutation.
The sign of a product of permutations is the product of their signs (sgn(αβ) = sgn(α)sgn(β)).
The product of two odd permutations is always an even permutation.
A permutation is in the alternating group A_n if and only if it is an even permutation.

Answer: Let τ be a fixed odd permutation in S₁₀. We need to show that any arbitrary odd permutation σ in S₁₀ can be written as a product of τ and some permutation π in A₁₀, i.e., σ = τπ. Since τ is an odd permutation, its inverse τ⁻¹ is also an odd permutation. This is a fundamental property, as the sign of a permutation and its inverse are the same, i.e., sgn(τ⁻¹) = sgn(τ). Consider the permutation π = τ⁻¹σ. We determine the parity of π. We know that sgn(π) = sgn(τ⁻¹σ) = sgn(τ⁻¹) × sgn(σ). Since both...

Q3c. List two distinct cosets of <r > in D10, where r is a reflection in D10.

Key Points:

D10 is the dihedral group of order 10, comprising 5 rotations and 5 reflections.
A reflection 'σ' in D10 satisfies σ² = e, generating subgroup <σ> = {e, σ}.
Cosets of a subgroup H in G are sets gH = {gh | h ∈ H} for g ∈ G.
The number of distinct cosets of <σ> in D10 is |D10|/|<σ>| = 10/2 = 5.

Answer: The dihedral group D10 has 10 elements, comprising 5 rotations (including identity) and 5 reflections. Let 'σ' represent a specific reflection in D10. The subgroup generated by this reflection, denoted as <σ>, consists of the identity element 'e' and the reflection 'σ' itself, since σ² = e. Therefore, our subgroup H is H = <σ> = {e, σ}, which has an order of 2. To find distinct left cosets of H in D10, we use elements from D10 to multiply by H. The total number of distinct cosets is determined ...

Q3d. Give the smallest n " N_for which A₁ is non-abelian. Justify your answer.

Key Points:

A_n is the group of even permutations of n elements.
A group is abelian if all its elements commute.
A₁ and A₂ are abelian, containing only the identity.
A₃ is abelian, as it is a cyclic group of prime order 3.

Answer: The smallest positive integer n for which the alternating group A_n is non-abelian is n=4. A_n is the group of all even permutations of n elements. A group is defined as abelian if all its elements commute under the group operation. For n=1 and n=2, A₁ and A₂ contain only the identity element {id}, making them trivially abelian. For n=3, A₃ has order 3!/2 = 3. Any group of prime order is cyclic, and all cyclic groups are abelian, so A₃ is abelian. For n=4, A₄ has order 4!/2 = 12. To demonstrat...

Q4. Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma: Let H and K be subgroups of a group G and H' and K' be normal subgroups of H and K, respectively. Then

Key Points:

Zassenhaus Lemma establishes an isomorphism between two 'butterfly' quotient groups.
The proof fundamentally relies on the Fundamental Theorem of Homomorphism (FTH) and the Second Isomorphism Theorem.
A crucial step is proving (H' ∩ K)(H ∩ K') is a normal subgroup of (H ∩ K).
Normality of H'(H ∩ K') in H'(H ∩ K) is established through subgroup properties.

Answer: The Zassenhaus Lemma, also known as the Butterfly Lemma, is a fundamental result in group theory that describes the relationship between subgroups and their normal subgroups. It is elegantly proven using the Fundamental Theorem of Homomorphism (FTH) and its direct consequence, the Second Isomorphism Theorem. This lemma is often applied in proving the Jordan-Hölder theorem, illustrating the 'butterfly' structure of certain lattices of subgroups. Let G be a group. Let H and K be subgroups of G, a...

Q4. Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma: Let H and K be subgroups of a group G and H' and K' be normal subgroups of H and K, respectively. Then

i)) H'(H∩K´)<H′(HK) (375 words)
ii)) K'(H'ƆK)<K'(H∩K) (375 words)

Key Points:

Zassenhaus Lemma establishes an isomorphism between two 'butterfly' quotient groups.
The proof fundamentally relies on the Fundamental Theorem of Homomorphism (FTH) and the Second Isomorphism Theorem.
A crucial step is proving (H' ∩ K)(H ∩ K') is a normal subgroup of (H ∩ K).
Normality of H'(H ∩ K') in H'(H ∩ K) is established through subgroup properties.

Q5. Which of the following statements are true, and which are false? Give reasons for your answers.

i)) For any ring R and a, b " R, (a+b)² = a² + 2ab + b². (100 words)
ii)) Every ring has at least two elements. (100 words)
iii)) If R is a ring with identity and I is an ideal of R, then the identity of R/I is the same as the identity of R. (100 words)
iv)) If f : R → S is a ring homomorphism, then it is a group homomorphism from (R, +) to (S, +). (100 words)
v)) If R is a ring, then any ring homomorphism from R×R_into Ris surjective. (100 words)

Key Points:

The identity `(a+b)² = a² + 2ab + b²` is only true in commutative rings, where `ab = ba`.
The trivial ring `{0}` is a valid ring with only one element, making the statement false.
The identity of `R/I` is `1+I`, which is a coset, distinct from `1 ∈ R` unless `I` is the zero ideal.
A ring homomorphism `f: R → S` preserves addition `f(a+b)=f(a)+f(b)`, fulfilling the group homomorphism definition for `(R,+)` and `(S,+)`.

Answer: As an expert IGNOU tutor, I will now evaluate the given statements regarding ring theory, providing detailed explanations and reasons based on the fundamental definitions and properties typically covered in the BMTC-134 Algebra course. Each sub-question will address its specific truth value and provide a concise justification, respecting the individual word limits.

Q6a. For an ideal I of a commutative ring R, define √I = {x R|x" I for some n N}. Show that

i)) √I is an ideal of R. (83 words)
ii)) I S √I. (83 words)
iii)) I ≠ √I in some cases. (83 words)

Key Points:

√I is the set of elements whose powers belong to ideal I.
To prove √I is an ideal, verify closure under subtraction and multiplication by ring elements.
I ⊆ √I because every x ∈ I implies x¹ ∈ I, satisfying the radical definition.
I ≠ √I when elements not in I have powers in I, e.g., 2 ∈ √<8> but 2 ∉ <8> in Z.

Answer: The radical of an ideal, denoted √I, is a fundamental concept in commutative ring theory, representing the set of all elements whose powers eventually fall into the ideal I. It plays a crucial role in understanding prime ideals and the nilradical of a ring, as discussed in block 3 of BMTC-134 Algebra. Demonstrating that √I is itself an ideal, that it always contains the original ideal I, and providing cases where it strictly differs from I are essential for grasping its properties.

Q6b. Is R/I × R/J ~ R/(I×J) for any two ideals I and J of a ring R? Give reasons for your answer.

Key Points:

The isomorphism R/I × R/J ~ R/(I×J) is generally false.
It holds true only if ideals I and J are comaximal (I + J = R).
Chinese Remainder Theorem (CRT) states R/(I ∩ J) ~ R/I × R/J if I + J = R.
If I and J are comaximal, then their product IJ equals their intersection I ∩ J.

Answer: No, the statement R/I × R/J ~ R/(I×J) is not generally true for any two ideals I and J of a ring R. This isomorphism only holds under a specific condition: when the ideals I and J are *comaximal*. This concept is fundamental in ring theory, often covered in BMTC-134, Unit 9, regarding the Chinese Remainder Theorem (CRT). According to the Chinese Remainder Theorem for rings, if R is a commutative ring with unity and I, J are ideals such that I + J = R (meaning they are comaximal), then R/(I ∩ J)...

Q7. Let S be a set, Raring and f be a 1-1 mapping of S onto R. Define + and on S by: x + y = f¯¹(f(x))+f(y)); x• y = f¯¹(f(x)f(y)); ∀ x, y S. Show that (S, +, ·) is a ring isomorphic to R.

Key Points:

To show (S, +, ·) is a ring, verify abelian group for addition, semigroup for multiplication, and distributivity.
Operations on S are defined via f and f⁻¹, leveraging all ring properties from R.
Additive identity in S is f⁻¹(0_R); additive inverse of x in S is f⁻¹(-f(x)).
A ring isomorphism is a bijective map preserving both addition and multiplication.

Answer: To demonstrate that (S, +, ·) is a ring isomorphic to R, we must establish two primary points: first, that (S, +, ·) satisfies all the ring axioms, and second, that the given map f is a ring isomorphism. This construction effectively 'transfers' the ring structure from R to S using the bijection f and its inverse f⁻¹. ### Proving (S, +, ·) is a Ring **1. (S, +) is an Abelian Group:** **Closure and Associativity:** For x, y ∈ S, x + y = f⁻¹(f(x) + f(y)) ∈ S because f(x)+f(y) ∈ R (closure in R)...

Q8. Which of the following statements are true, and which are false? Give reasons for your answers.

i)) If k is a field, then so is k×k. (100 words)
ii)) If R is an integral domain and I is an ideal of R, then Char (R) = Char (R/I). (100 words)
iii)) In a domain, every prime ideal is a maximal ideal. (100 words)
iv)) If R is a ring with zero divisors, and S is a subring of R, then S has zero divisors. (100 words)
v)) If R is a ring and f(x)" R[x] is of degree n"N, then f(x) has exactly n roots in R. (100 words)

Key Points:

Direct product k×k is not a field if k has >1 element due to zero divisors.
Characteristic of an integral domain R is not always equal to Char(R/I).
Not all prime ideals in a domain are maximal; counterexample: ⟨x⟩ in Z[x].
Subrings do not always inherit zero divisors from the larger ring (e.g., {0,3} in Z₆).

Answer: This response evaluates five statements regarding fundamental concepts in ring theory and field theory, as typically covered in the BMTC-134 Algebra course. Each statement is analyzed for its truthfulness, and detailed reasons are provided, often with counterexamples, to support the conclusion. The analysis draws upon definitions and theorems related to fields, integral domains, rings, ideals, characteristics of rings, and polynomial roots, emphasizing key distinctions that are crucial for under...

Q9a. Find all the units of Z[√-7].

Key Points:

Units in an integral domain are elements with multiplicative inverses within the domain.
For Z[√-d], the norm of an element a + b√-d is N(a + b√-d) = a² + db².
An element in Z[√-d] is a unit if and only if its norm N(α) is equal to ±1.
For Z[√-7], the condition for units simplifies to solving the Diophantine equation a² + 7b² = 1.

Answer: As discussed in the BMTC-134 course material, specifically in units covering integral domains and number rings, an element 'u' in an integral domain 'R' is defined as a unit if there exists an element 'v' in 'R' such that uv = 1. For rings of the form Z[√d] or Z[√-d], like Z[√-7], the concept of a norm is essential for identifying these units. For an element α = a + b√-d, where a and b are integers, the norm is defined as N(α) = N(a + b√-d) = a² + d(-b²). In our case, for Z[√-7], d = 7, so the ...

Q9b. Check whether or not Z[x]/<8x³ + 6x²-9x+24> is a field.

Key Points:

A quotient ring R/I is a field iff I is a maximal ideal in R.
Z[x] is the ring of polynomials with integer coefficients.
For a non-constant polynomial f(x), <f(x)> is never a maximal ideal in Z[x].
Maximal ideals in Z[x] are of the form <p, g(x)>, not solely principal.

Answer: To determine if Z[x]/<8x³ + 6x²-9x+24> is a field, we refer to a fundamental theorem from Unit 3/4 of BMTC-134: A quotient ring R/I is a field if and only if R is a commutative ring with unity and I is a maximal ideal in R. Here, R = Z[x] and I = <8x³ + 6x²-9x+24>. Let f(x) = 8x³ + 6x² - 9x + 24. For Z[x]/<f(x)> to be a field, the ideal <f(x)> must be maximal in Z[x]. However, a principal ideal generated by a non-constant polynomial in Z[x] is never maximal. This is because Z[x] is not a Princi...

Q9c. Construct a field with 125 elements.

Key Points:

Finite fields have pⁿ elements, where p is prime and n ≥ 1.
For 125 elements, the field is GF(5³), with p=5 and n=3.
GF(pⁿ) is constructed as Z_p[x] / <f(x)>, where f(x) is irreducible.
f(x) must be an irreducible polynomial of degree n over Z_p.

Answer: To construct a field with 125 elements, we first recognize that the number of elements in a finite field must be of the form pⁿ, where p is a prime number and n is a positive integer. For 125, we identify 125 = 5³, which means p = 5 and n = 3. This implies we need to construct a finite field of order 5³, denoted as GF(5³). As per the course material (e.g., Block 4, Unit 16), such a field can be constructed as the quotient ring Z₅[x] / <f(x)>, where Z₅[x] is the ring of polynomials with coeffici...

BMTC-134 Solved Assignment — January 2026 / July 2026

ALGEBRA | IGNOU Bachelor of Science (General) (CBCS) (BSCG)

BMTC-134—January 2026 / July 2026

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BMTC-134 Assignment Questions — January 2026

Q1. Which of the following statements are true? Give reasons for your answers.

Q2a. Prove that every non-trivial subgroup of a cyclic group has finite index. Hence prove that (Q, +) is not cyclic.

Q2b. Let G be an infinite group such that for any non-trivial subgroup H of G, |G : H| < ". Then prove that

Q2c. Prove that a cyclic group with only one generator can have at most 2 elements.

Q3a. Using Cayley's theorem, find the permutation group to which a cyclic group of order 12 is isomorphic.

Q3b. Let τ be a fixed odd permutation in S10. Show that every odd permutation in S10 is a product of τ and some permutation in A10.

Q3c. List two distinct cosets of <r > in D10, where r is a reflection in D10.

Q3d. Give the smallest n " N_for which A₁ is non-abelian. Justify your answer.

Q4. Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma: Let H and K be subgroups of a group G and H' and K' be normal subgroups of H and K, respectively. Then

Q4. Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma: Let H and K be subgroups of a group G and H' and K' be normal subgroups of H and K, respectively. Then

Q5. Which of the following statements are true, and which are false? Give reasons for your answers.

Q6a. For an ideal I of a commutative ring R, define √I = {x R|x" I for some n N}. Show that

Q6b. Is R/I × R/J ~ R/(I×J) for any two ideals I and J of a ring R? Give reasons for your answer.

Q7. Let S be a set, Raring and f be a 1-1 mapping of S onto R. Define + and on S by: x + y = f¯¹(f(x))+f(y)); x• y = f¯¹(f(x)f(y)); ∀ x, y S. Show that (S, +, ·) is a ring isomorphic to R.

Q8. Which of the following statements are true, and which are false? Give reasons for your answers.

Q9a. Find all the units of Z[√-7].

Q9b. Check whether or not Z[x]/<8x³ + 6x²-9x+24> is a field.

Q9c. Construct a field with 125 elements.

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