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Q1. (a) Define the Euler φ-function. Further, find φ(8) and 7 ^φ(8) (mod 8). 3 (b) List all the generators of the group Z₁₅. Justify your list. 2 (c) Find all the Sylow subgroups of a group of order 35. Is this group cyclic? Justify your answer. 5

Ans. (a) Definition of Euler φ-function: For a positive integer n, the Euler’s totient function, denoted by φ(n), counts the number of positive integers up to a given integer n that are relatively prime to n. That is, φ(n) = |{k ∈ {1, 2, …, n} | gcd(k, n) = 1}|. Finding φ(8): We need to find the integers less than or equal to 8 that are relatively prime to 8. The integers are {1, 2, 3, 4, 5, 6, 7, 8}. The numbers among these that are relatively prime to 8 (i.e., their greatest common divisor with 8 is 1) are {1, 3, 5, 7}. Thus, φ(8) = 4 . Finding 7 ^φ(8) (mod 8): We need to compute 7⁴ (mod 8). By Euler’s Totient Theorem, if gcd(a, n) = 1, then a ^φ(n) ≡ 1 (mod n). Here, a = 7 and n = 8. Since gcd(7, 8) = 1, the theorem applies. 7 ^φ(8) ≡ 1 (mod 8) 7⁴ ≡ 1 (mod 8) Therefore, the value of 7 ^φ(8) (mod 8) is 1 .

(b) The group Z₁₅ is the cyclic group of integers modulo 15 under addition. The generators of this group are the elements [k] ∈ Z₁₅ such that gcd(k, 15) = 1 . We need to find the values of k in the set {1, 2, …, 14}. Since 15 = 3 × 5, k must not be a multiple of 3 or 5. The positive integers less than 15 that are not divisible by 3 or 5 are: 1, 2, 4, 7, 8, 11, 13, 14. Thus, the list of all generators of Z₁₅ is: [1], [2], [4], [7], [8], [11], [13], [14] . The total number of generators is φ(15) = φ(3)φ(5) = (3-1)(5-1) = 2 × 4 = 8, which matches our list.

(c) Let G be a group of order 35. So, |G| = 35 = 5 × 7. Sylow Subgroups:

• Sylow 5-subgroups: Let n₅ be the number of Sylow 5-subgroups. By Sylow’s Third Theorem, n₅ must divide 7 and n₅ ≡ 1 (mod 5). The divisors of 7 are 1 and 7. Checking the congruence: 1 ≡ 1 (mod 5) and 7 ≡ 2 (mod 5). The only possibility is n₅ = 1 . This means there is a unique Sylow 5-subgroup, let’s call it P. Since it is unique, it is a normal subgroup of G, and |P|=5.
• Sylow 7-subgroups: Let n₇ be the number of Sylow 7-subgroups. By Sylow’s Third Theorem, n₇ must divide 5 and n₇ ≡ 1 (mod 7). The divisors of 5 are 1 and 5. Checking the congruence: 1 ≡ 1 (mod 7) and 5 ≡ 5 (mod 7). The only possibility is n₇ = 1 . This means there is a unique Sylow 7-subgroup, let’s call it Q. Since it is unique, it is a normal subgroup of G, and |Q|=7.

Is the group cyclic?

Yes, the group is cyclic. The justification is as follows:

1. P and Q are groups of prime order 5 and 7, respectively. Therefore, both P and Q are cyclic groups.
2. P and Q are both normal subgroups of G.
3. Since |P| and |Q| are coprime, their intersection is the trivial subgroup: P ∩ Q = {e}.
4. The number of elements in PQ is |PQ| = (|P| |Q|) / |P ∩ Q| = (5 × 7) / 1 = 35 = |G|. So, G = PQ.
5. Since P and Q are normal subgroups of G, G is isomorphic to their internal direct product, G ≅ P × Q.
6. Since P ≅ Z₅ and Q ≅ Z₇, we have G ≅ Z₅ × Z₇.
7. Because gcd(5, 7) = 1, the direct product Z₅ × Z₇ is isomorphic to the cyclic group Z₅ₓ₇ = Z₃₅.

Therefore, any group of order 35 is

cyclic

.

Q2. (a) Let G be the set of all real numbers except -1. Define an operation on G by a b = ab + a + b for all a, b ∈ G. Check whether or not G is an abelian group. 4 (b) Let R be a ring such that r² = r for all r ∈ R. Show that R is commutative. Also find the characteristic of R. 3 (c) Give an example, with justification, of a ring R which is not an integral domain, but for which every ideal is principal. 3

Ans. (a) To check if (G, ), where G = ℝ \ {-1} and a b = ab + a + b, is an abelian group, we verify the group axioms:

1. Closure: Let a, b ∈ G, so a ≠ -1 and b ≠ -1. We must show a b ∈ G, i.e., a b ≠ -1. Assume a b = -1. Then ab + a + b = -1 ⇒ ab + a + b + 1 = 0 ⇒ a(b+1) + 1(b+1) = 0 ⇒ (a+1)(b+1) = 0. This implies a+1=0 or b+1=0, so a=-1 or b=-1. This contradicts that a, b ∈ G. Thus, a b ≠ -1, so G is closed under *.
2. Associativity: For a, b, c ∈ G: (a b) c = (ab+a+b)*c = (ab+a+b)c + (ab+a+b) + c = abc+ac+bc+ab+a+b+c. a (b c) = a*(bc+b+c) = a(bc+b+c) + a + (bc+b+c) = abc+ab+ac+a+bc+b+c. The results are equal, so * is associative.
3. Identity Element: We seek an element e ∈ G such that a*e = a for all a ∈ G. ae + a + e = a ⇒ ae + e = 0 ⇒ e(a+1) = 0. Since this must hold for all a ∈ G (a ≠ -1), we must have e = 0 . Since 0 ≠ -1, 0 ∈ G. Thus, 0 is the identity element.
4. Inverse Element: For each a ∈ G, we seek an a’ ∈ G such that a*a’ = 0. aa’ + a + a’ = 0 ⇒ a'(a+1) = -a ⇒ a’ = -a/(a+1). Since a ≠ -1, the inverse a’ is well-defined. We must also check a’ ≠ -1. If a’ = -1, then -a/(a+1) = -1 ⇒ -a = -a-1 ⇒ 0 = -1, a contradiction. So a’ ∈ G. Every element has an inverse.
5. Commutativity: We check if a b = b a. a*b = ab + a + b. b*a = ba + b + a. Since multiplication and addition of real numbers are commutative, ab = ba and a+b = b+a. Hence, a b = b a.

Since all axioms are satisfied,

G is an abelian group

.

(b) Given R is a ring where r² = r for all r ∈ R. Such a ring is a Boolean ring. Commutativity: Let a, b ∈ R. Then (a+b) ∈ R, so (a+b)² = a+b. (a+b)(a+b) = a+b ⇒ a² + ab + ba + b² = a+b. Since a²=a and b²=b, this becomes a + ab + ba + b = a+b. Subtracting a+b from both sides gives ab + ba = 0, which means ab = -ba . Now for any r ∈ R, consider r+r. (r+r)² = r+r. (r+r)² = r² + r² + r² + r² = r+r+r+r = 4r. Note that (r+r) is 2r, so (2r)² = 4r² = 4r. Thus, 4r = 2r, which implies 2r = 0 for all r ∈ R. This means r = -r for all r ∈ R. Substituting this into our earlier result ab = -ba, we get ab = ba. Thus, R is commutative . Characteristic: As shown above, 2r = 0 for all r ∈ R. If R is not the zero ring {0}, there exists a non-zero r, so the characteristic must be 2. If R is the zero ring, the characteristic is 1. For any non-trivial Boolean ring, the characteristic is 2 .

(c) We need an example of a ring that is not an integral domain (i.e., has zero divisors) but is a Principal Ideal Ring (PIR, a ring where every ideal is principal). Consider the ring Z₆ (the integers modulo 6).

• Not an Integral Domain: In Z₆, we have [2] ≠ [0] and [3] ≠ [0], but their product is [2] ⋅ [3] = [6] = [0]. Since Z₆ has zero divisors, it is not an integral domain.
• Every Ideal is Principal: The ideals of Zₙ correspond to the additive subgroups. Since (Z₆, +) is a cyclic group, all of its subgroups are also cyclic, and hence generated by a single element. The ideals of Z₆ are:
  • <[0]> = {[0]}
  • <[1]> = Z₆
  • <[2]> = {[0], [2], [4]}
  • <[3]> = {[0], [3]}

All these ideals are generated by a single element, so they are principal ideals.

Therefore,

Z₆

is a ring which is not an integral domain, but for which every ideal is principal. Another simple example is Z₄.

Q3. (a) If N is a subgroup of index 2 in a group G, then show that N is a normal subgroup in G. 3 (b) Let f:Z→Z be defined by f(x)=x²-2. Find f⁻¹({0, 1, 2}). 2 (c) Check whether or not Q[x] / <x² + 2x + 2> is a field. 5

Ans. (a) Given that N is a subgroup of G with index [G:N] = 2. This means there are exactly two distinct left cosets of N in G, and also two distinct right cosets. One of the left cosets is N itself (since eN = N). Let the other left coset be gN for some g ∈ G, g ∉ N. Then G is the disjoint union G = N ∪ gN. Similarly, one of the right cosets is N (since Ne = N). The other must be Ng. Then G is the disjoint union G = N ∪ Ng. We need to show that N is normal, which means we must show that gN = Ng for all g ∈ G.

• Case 1: g ∈ N. If g is in N, then gN = N (by closure property of a subgroup) and Ng = N. Thus, gN = Ng.
• Case 2: g ∉ N. If g is not in N, then the left coset gN is not equal to N. Since there are only two left cosets, gN must be the other one. So, gN = G \ N. Similarly, the right coset Ng is not equal to N. It must be the other right coset. So, Ng = G \ N. Therefore, gN = Ng.

Since gN = Ng for all g ∈ G,

N is a normal subgroup in G

.

(b) The function is f: Z → Z, defined by f(x) = x² – 2. We need to find f⁻¹({0, 1, 2}). By definition, f⁻¹({0, 1, 2}) is the set of all integers x such that f(x) is in the set {0, 1, 2}. We solve for x in each case:

1. f(x) = 0: x² – 2 = 0 ⇒ x² = 2. There are no integer solutions for x.
2. f(x) = 1: x² – 2 = 1 ⇒ x² = 3. There are no integer solutions for x.
3. f(x) = 2: x² – 2 = 2 ⇒ x² = 4. The integer solutions are x = 2 and x = -2.

Combining the solutions from all cases, we find the set of pre-images.

Therefore,

f⁻¹({0, 1, 2}) = {-2, 2}

.

(c) A quotient ring R/I is a field if and only if the ideal I is a maximal ideal in the ring R. Here, the ring is R = Q[x], the ring of polynomials with rational coefficients. This is a Principal Ideal Domain (PID). In a PID, a non-zero ideal <p(x)> is maximal if and only if the polynomial p(x) is irreducible over the base field. The base field is Q (the rational numbers) and the polynomial is p(x) = x² + 2x + 2. To check for irreducibility, we can find the roots of p(x) using the quadratic formula: x = [-b ± √(b² – 4ac)] / 2a. x = [-2 ± √(2² – 4⋅1⋅2)] / 2 = [-2 ± √(4 – 8)] / 2 = [-2 ± √(-4)] / 2 = [-2 ± 2i] / 2 = -1 ± i. The roots of the polynomial are -1 + i and -1 – i. These are complex numbers, not rational numbers. Since the quadratic polynomial p(x) has no roots in the field Q, it is irreducible over Q. Since p(x) = x² + 2x + 2 is irreducible over Q, the ideal I = <x² + 2x + 2> is a maximal ideal in Q[x]. Therefore, the quotient ring Q[x] / <x² + 2x + 2> is a field .

Q4. (a) Express the permutation σ = (1 2 3 4 5 6 7 8 9 / 8 9 2 3 7 6 5 4 1) as a product of disjoint cycles. Is σ even? Give reasons for your answer. 3 (b) State the Fundamental Theorem of Homomorphism for rings. Further, apply it to prove that R[x]/<x> ≅ R. 7

Ans. (a) The permutation is given in two-line notation: σ = (1 2 3 4 5 6 7 8 9 / 8 9 2 3 7 6 5 4 1) Product of disjoint cycles: We trace the paths of the elements: 1 maps to 8, 8 maps to 4, 4 maps to 3, 3 maps to 2, 2 maps to 9, and 9 maps to 1. This forms the cycle (1 8 4 3 2 9). 5 maps to 7, and 7 maps to 5. This forms the cycle (5 7). 6 maps to 6, which is a fixed point and usually omitted in cycle notation. So, the disjoint cycle decomposition is σ = (1 8 4 3 2 9)(5 7) .

Is σ even? A permutation is even if it can be expressed as a product of an even number of transpositions (2-cycles). Method 1: Decomposing into transpositions. A k-cycle can be written as a product of k-1 transpositions.

• The 6-cycle (1 8 4 3 2 9) = (1 9)(1 2)(1 3)(1 4)(1 8). This is 5 transpositions (odd).
• The 2-cycle (5 7) is 1 transposition (odd).

The total number of transpositions is 5 + 1 = 6.

Since 6 is an even number,

σ is an even permutation

.

Method 2: Parity of cycles.

The parity of a k-cycle is odd if k is even, and even if k is odd.

• (1 8 4 3 2 9) is a 6-cycle (length 6, even), so it is an odd permutation.
• (5 7) is a 2-cycle (length 2, even), so it is an odd permutation.

The permutation σ is the product of two odd permutations. The product of two odd permutations is an even permutation. Thus,

σ is even

.

(b) Fundamental Theorem of Homomorphism for Rings: Let R and S be rings and let f: R → S be a surjective ring homomorphism. Then the quotient ring R/ker(f) is isomorphic to the ring S. In symbols, R/ker(f) ≅ S .

Proof of R[x]/<x> ≅ R: To prove this, we will define a suitable surjective homomorphism from R[x] to R and then apply the theorem. Consider the evaluation map f: R[x] → R defined by f(p(x)) = p(0) .

1. Show f is a ring homomorphism: Let p(x), q(x) ∈ R[x].
   • f(p(x) + q(x)) = (p+q)(0) = p(0) + q(0) = f(p(x)) + f(q(x)).
   • f(p(x)q(x)) = (pq)(0) = p(0)q(0) = f(p(x))f(q(x)).

   Thus, f is a ring homomorphism.

2. Show f is surjective: Let r be any element in the codomain R. We need to find a polynomial p(x) ∈ R[x] such that f(p(x)) = r. Consider the constant polynomial p(x) = r. This polynomial is in R[x]. Then f(p(x)) = f(r) = p(0) = r. Since for every r ∈ R, there exists a pre-image in R[x], f is surjective.
3. Find the kernel of f: The kernel is ker(f) = {p(x) ∈ R[x] | f(p(x)) = 0}. f(p(x)) = p(0) = 0. If p(x) = a₀ + a₁x + … + aₙxⁿ, then p(0) = a₀. So p(0) = 0 implies the constant term a₀ is 0. Any such polynomial can be written as p(x) = a₁x + a₂x² + … + aₙxⁿ = x(a₁ + a₂x + … + aₙxⁿ⁻¹). This means p(x) is a multiple of x. The set of all polynomials that are multiples of x is precisely the principal ideal generated by x. Thus, ker(f) = <x> .
4. Apply the Fundamental Theorem: We have a surjective ring homomorphism f: R[x] → R with kernel <x>. By the Fundamental Theorem of Homomorphism, R[x]/ker(f) ≅ R. Substituting the kernel, we get R[x]/<x> ≅ R .

Q5. (a) Give an example, with justification, to show that if N is a normal subgroup of a group G, then the centre of N may not be contained in the centre of G. 3 (b) In a Euclidean domain R, show that for any non-zero r ∈ R, d(1) ≤ d(r), where d is a Euclidean valuation. 2 (c) Give an example, with justification, of a ring with identity that has a subring not containing identity. 2 (d) Use the principle of mathematical induction to prove that n² < 2ⁿ for all n ∈ N, n > 3. 3

Ans. (a) We need an example of a group G and a normal subgroup N ⊲ G such that the center of N, Z(N), is not a subset of the center of G, Z(G). Consider the symmetric group on 3 elements, G = S₃ . |S₃| = 6. G = {e, (12), (13), (23), (123), (132)}. Let N = A₃ = {e, (123), (132)} , the subgroup of even permutations. N is a normal subgroup of G because its index is [G:N] = 6/3 = 2. Now we find the centers:

• Center of N, Z(N): N is a cyclic group of order 3, so it is abelian. Therefore, its center is the entire group N. Z(N) = N = {e, (123), (132)} .
• Center of G, Z(G): The center of S₃ is the trivial subgroup {e}. For example, (123)(12) = (13) but (12)(123) = (23), so (123) does not commute with (12). No non-identity element commutes with all elements of S₃. Thus, Z(G) = {e} .

Comparing the two centers, Z(N) = {e, (123), (132)} and Z(G) = {e}.

Clearly,

Z(N) is not contained in Z(G)

, because (123) ∈ Z(N) but (123) ∉ Z(G). This serves as the required example.

(b) Let R be a Euclidean domain with a Euclidean valuation d. One of the properties of the Euclidean valuation d is that for any two non-zero elements a, b ∈ R, d(a) ≤ d(ab) . Let r be any non-zero element in R. The multiplicative identity, 1, is also a non-zero element in R. We can write r as r = 1 ⋅ r. Now, apply the property d(a) ≤ d(ab) by setting a = 1 and b = r. Since both are non-zero, the property holds. We get: d(1) ≤ d(1 ⋅ r) Which simplifies to d(1) ≤ d(r) . This shows that the d-value of the identity element 1 is less than or equal to the d-value of any other non-zero element in the domain.

(c) We need an example of a ring R with identity, which has a subring S that does not contain the identity of R. Let the ring be R = Z₆ = {[0], [1], [2], [3], [4], [5]} . The identity element of R is [1]. Consider the subset S = {[0], [2], [4]} . We check if S is a subring of R using the subring test:

1. S is non-empty.
2. S is closed under subtraction: [2]-[4] = [-2]=[4]∈S, [2]-[2]=[0]∈S, etc. It forms a subgroup under addition.
3. S is closed under multiplication: [2]⋅[2]=[4]∈S, [2]⋅[4]=[8]=[2]∈S, [4]⋅[4]=[16]=[4]∈S.

Since S is closed under subtraction and multiplication, it is a subring of R.

The identity element of the ring R is [1].

Clearly,

[1] ∉ S

.

Therefore, S is a subring of Z₆ that does not contain the identity of Z₆.

(

Note: S has its own multiplicative identity, which is [4], since [2]⋅[4]=[2] and [4]⋅[4]=[4]. But this is not the identity of the parent ring R.

)

(d) We want to prove n² < 2ⁿ for all n ∈ N, n > 3. Note on the question: The statement is false for n=4, since 4² = 16 and 2⁴ = 16, so 4² < 2⁴ is not true. The inequality holds for n ≥ 5. We will prove it for n ≥ 5, acknowledging the error in the question’s range. Let P(n) be the statement n² < 2ⁿ. Base Case: For n = 5. LHS = 5² = 25. RHS = 2⁵ = 32. Since 25 < 32, P(5) is true. Inductive Hypothesis: Assume P(k) is true for some integer k ≥ 5. That is, assume k² < 2ᵏ . Inductive Step: We must prove that P(k+1) is true, i.e., (k+1)² < 2ᵏ⁺¹. Start with the LHS: (k+1)² = k² + 2k + 1. By the inductive hypothesis, k² < 2ᵏ. So, (k+1)² < 2ᵏ + 2k + 1. Now we need to show that 2ᵏ + 2k + 1 < 2ᵏ⁺¹. We know 2ᵏ⁺¹ = 2 ⋅ 2ᵏ = 2ᵏ + 2ᵏ. So we need to show that 2k + 1 < 2ᵏ for k ≥ 5. This inequality can be separately proven by induction for k ≥ 3. For k=5, 2(5)+1 = 11 and 2⁵=32. Clearly 11 < 32. Since 2k+1 < 2ᵏ for k≥5, we can conclude: (k+1)² < 2ᵏ + 2k + 1 < 2ᵏ + 2ᵏ = 2ᵏ⁺¹. Thus, (k+1)² < 2ᵏ⁺¹, and P(k+1) is true. By the principle of mathematical induction, n² < 2ⁿ is true for all integers n ≥ 5 .

Q6. (a) Describe all subgroups of (Z, +) which contain 5Z and 7Z. 2 (b) Let T be the set of straight lines in a plane. Define a relation on T by L₁RL₂ iff L₁ is perpendicular to L₂. Is the relation reflexive? Is it symmetric? Is it transitive? Justify your answers. 3 (c) Let R = {a/b | a,b∈Z s.t. b=2ᵐ·3ⁿ, m,n ∈ Z, m, n≥0}. Show that R is a ring with respect to the usual addition and multiplication of rational numbers. Is R commutative? Does R have an identity element? Justify your answers. 5

Ans. (a) The subgroups of (Z, +) are all of the form nZ for some non-negative integer n. We are looking for subgroups nZ that contain both 5Z and 7Z.

• If 5Z ⊆ nZ , it means every multiple of 5 is also a multiple of n. This is only possible if n divides 5. So, n must be 1 or 5.
• If 7Z ⊆ nZ , it means every multiple of 7 is also a multiple of n. This is only possible if n divides 7. So, n must be 1 or 7.

For both conditions to hold, n must be a common divisor of 5 and 7. Since 5 and 7 are coprime, their only positive common divisor is 1.

So, n = 1.

The only subgroup containing both 5Z and 7Z is

1Z = Z

.

Alternatively, the smallest subgroup containing two subgroups H and K is their sum H+K. Here, H=5Z and K=7Z. The subgroup generated by them is 5Z + 7Z = {5x + 7y | x, y ∈ Z}. Since gcd(5, 7) = 1, this subgroup is 1Z = Z. Any other subgroup containing 5Z and 7Z must contain Z, and must therefore be Z itself.

(b) The relation R is defined on the set T of lines in a plane by L₁ R L₂ ⇔ L₁ is perpendicular to L₂.

• Reflexive: Is L R L true for any line L? This means, is a line perpendicular to itself? No. A line makes an angle of 0° with itself, not 90°. Therefore, the relation is not reflexive .
• Symmetric: If L₁ R L₂, is L₂ R L₁ true? If L₁ is perpendicular to L₂, then L₂ is also perpendicular to L₁. The property of being perpendicular is mutual. Therefore, the relation is symmetric .
• Transitive: If L₁ R L₂ and L₂ R L₃, is L₁ R L₃ true? Let L₁ be the x-axis. If L₁ ⊥ L₂, then L₂ must be a vertical line (e.g., the y-axis). If L₂ ⊥ L₃, then L₃ must be a horizontal line, parallel to the x-axis. In this case, L₁ is parallel to L₃, not perpendicular to it. Therefore, the relation is not transitive .

(c) Given R = {a/b | a,b∈Z, b=2ᵐ3ⁿ, m, n≥0}. We use the subring test to show it is a ring. R is a subset of the rational numbers Q.

1. R is non-empty, as 0 = 0/1 ∈ R (with m=0, n=0).
2. Closure under subtraction: Let x = a/b and y = c/d be in R, where b=2ᵐ3ⁿ and d=2ᵖ3۹. x – y = a/b – c/d = (ad – bc) / bd. The numerator ad-bc is an integer. The denominator is bd = (2ᵐ3ⁿ)(2ᵖ3۹) = 2ᵐ⁺ᵖ3ⁿ⁺۹. Since m,n,p,q are non-negative, so are m+p and n+q. Thus the denominator is of the required form. Hence, x-y ∈ R.
3. Closure under multiplication: x ⋅ y = (a/b) ⋅ (c/d) = ac / bd. The numerator ac is an integer. The denominator bd = 2ᵐ⁺ᵖ3ⁿ⁺۹ is of the required form. Hence, x⋅y ∈ R.

By the subring test,

R is a subring of Q, and therefore is a ring

.

• Is R commutative? Yes. Since R is a subring of the field of rational numbers Q, which is commutative, the multiplication in R must also be commutative .
• Does R have an identity element? Yes. The multiplicative identity in Q is 1. We can write 1 as 1/1. The denominator b=1 can be expressed as 1 = 2⁰3⁰, which fits the required form with m=0, n=0. Thus, 1 ∈ R. So, R has an identity element .

Q7. Which of the following statements are true and which are false? Give reason for your answers. (Marks will only be given for proper justification): 10 (i) In the Klein 4-group, every element has order 2. (ii) Z₁₈ and Z₃ × Z₆ are isomorphic groups. (iii) Z₉ has no non-zero nilpotent elements. (iv) In a ring with identity, a unit cannot be a zero divisor. (v) Zₚ is a field, where p is a prime.

Ans. (i) False. The Klein 4-group is V = {e, a, b, c}, where e is the identity. The non-identity elements a, b, and c each have order 2. However, the identity element e has order 1 . Since the statement says “every element”, it is false because of the identity element.

(ii) False. The group Z₁₈ is a cyclic group of order 18. The group Z₃ × Z₆ has order 3 × 6 = 18. However, a direct product Zₘ × Zₙ is cyclic if and only if gcd(m, n) = 1. Here, gcd(3, 6) = 3 ≠ 1 , so Z₃ × Z₆ is not a cyclic group. Since Z₁₈ is cyclic and Z₃ × Z₆ is not, they cannot be isomorphic.

(iii) False. A non-zero element ‘a’ in a ring is nilpotent if aⁿ = 0 for some positive integer n. In the ring Z₉, consider the element [3] . It is non-zero. We have [3]² = [9] = [0] (mod 9). Since [3]² = [0], [3] is a non-zero nilpotent element in Z₉. Therefore, the statement is false.

(iv) True. Let R be a ring with identity 1 and let u be a unit in R. This means there exists an inverse u⁻¹ ∈ R such that uu⁻¹ = u⁻¹u = 1. Now, assume for contradiction that u is also a zero divisor. Then there exists a non-zero element v ∈ R such that uv = 0. Multiplying this equation on the left by u⁻¹, we get: u⁻¹(uv) = u⁻¹(0) ⇒ (u⁻¹u)v = 0 ⇒ 1v = 0 ⇒ v = 0. This contradicts our assumption that v was non-zero. Thus, a unit cannot be a zero divisor.

(v) True. For Zₚ to be a field, it must be a commutative ring with identity in which every non-zero element has a multiplicative inverse. Zₚ is a commutative ring with identity [1]. Let [a] be a non-zero element in Zₚ (so a 0 mod p). Since p is a prime number and a is not a multiple of p, it follows that gcd(a, p) = 1 . By the Extended Euclidean Algorithm (or Bezout’s Identity), there exist integers x and y such that ax + py = 1. Taking this equation modulo p, we get ax ≡ 1 (mod p). This means [a][x] = [1] in Zₚ, so [x] is the multiplicative inverse of [a]. Since every non-zero element is a unit, Zₚ is a field.