Explanation: In a cyclic subgroup, ord(a^k)=ord(a)/gcd(ord(a),k). Here ord(a)=6 and k=3, so ord(a³)=6/gcd(6,3)=6/3=2.
If ord(a)=10, then ord(a²) equals
A 5
B 10
C 2
D 1
Explanation: Use ord(a^k)=n/gcd(n,k) for finite order n. With n=10 and k=2, gcd(10,2)=2, so ord(a²)=10/2=5.
In a finite group, equation x²=e implies x has order
A 2 only
B 1 or 2
C 1 only
D 4 only
Explanation: If x²=e, then repeating gives x has order dividing 2. So either x=e (order 1) or x≠e with smallest power 2 (order 2).
In any group, (ab)⁻¹ equals
A a⁻¹b⁻¹
B (ab) itself
C b⁻¹a⁻¹
D a⁻¹b
Explanation: Multiply (ab)(b⁻¹a⁻¹)=a(bb⁻¹)a⁻¹=aea⁻¹=e and similarly on the other side. The inverse reverses order due to associativity.
In any group, solving ax=b gives x=
A a⁻¹b
B ba⁻¹
C ab⁻¹
D b⁻¹a
Explanation: Multiply ax=b on the left by a⁻¹: a⁻¹(ax)=(a⁻¹a)x=ex=x, so x=a⁻¹b. Existence of inverses guarantees unique solution.
In any group, solving xa=b gives x=
A a⁻¹b
B b⁻¹a
C ab⁻¹
D ba⁻¹
Explanation: Multiply xa=b on the right by a⁻¹: (xa)a⁻¹=x(aa⁻¹)=xe=x, so x=ba⁻¹. Right cancellation uses the inverse on the right.
In a group, equation a²=a implies
A a has order 2
B a is generator
C a=e
D group abelian
Explanation: From a²=a, multiply by a⁻¹ on the left to get a=e. In a group, idempotent elements must be identity because inverses allow cancellation.
Number of elements in ⟨a⟩ equals
A ord(a)
B |G| always
C index of ⟨a⟩
D number of cosets
Explanation: ⟨a⟩ contains all distinct powers of a until identity repeats. The count of these distinct elements is exactly the order of a by definition.
If H is subgroup of G, then H always contains
A all of G
B no inverses
C identity e
D no identity
Explanation: Any subgroup must itself be a group under the same operation, so it must contain an identity element. That identity must be the same as G’s identity.
If H is a subgroup, then for h in H, element h⁻¹ is
A not in H
B only in G
C depends on coset
D in H
Explanation: Subgroup condition requires closure under inverses. So whenever h belongs to H, the inverse h⁻¹ must also belong to H, ensuring H forms a group.
A nonempty subset H is subgroup if for all a,b in H
A ab in G only
B a+b in ℤ
C ab⁻¹ in H
D a=b always
Explanation: This is the one-step subgroup test. If H is nonempty and closed under ab⁻¹, then it automatically has identity and inverses and is closed under operation.
In a finite group, number of left cosets of H equals
A |H|
B [G:H]
C ord(H)²
D φ(|H|)
Explanation: The index [G:H] is defined as the number of distinct left cosets of H in G. For finite groups it equals |G|/|H|.
If |G|=30, then possible subgroup order is
A 10
B 7
C 11
D 16
Explanation: By Lagrange, subgroup order must divide |G|. Divisors of 30 include 1,2,3,5,6,10,15,30. So 10 is possible; 7,11,16 are not.
If |G| is prime p, then number of proper subgroups is
A p
B p−1
C 1
D many always
Explanation: For prime order p, subgroup orders must divide p, so only 1 and p are possible. Thus only the trivial subgroup {e} is proper.
If g has order 12, then g⁴ has order
A 4
B 6
C 12
D 3
Explanation: ord(g^k)=n/gcd(n,k). Here n=12, k=4, gcd(12,4)=4, so ord(g⁴)=12/4=3. This is standard in cyclic subgroups.
Two left cosets gH and kH are equal iff
A g⁻¹k in H
B gk in H
C g=k always
D H is normal
Explanation: gH=kH exactly when k belongs to gH, meaning k=gh for some h in H. Rearranging gives g⁻¹k=h in H, proving the criterion.
Two left cosets gH and kH are disjoint when
A g=k
B H={e}
C gH ≠ kH
D G abelian
Explanation: Left cosets form a partition of the group. So any two cosets are either identical or disjoint. If they are not equal, they share no elements.
In quotient group G/N, identity element is
A e only
B G
C {e}
D N
Explanation: In G/N, elements are cosets. The identity coset is eN, which equals N. Multiplying N with any coset gN returns gN, matching identity behavior.
In G/N, inverse of coset gN is
A Ng⁻¹
B g⁻¹N
C (gN)²
D N itself
Explanation: (gN)(g⁻¹N)=(gg⁻¹)N=eN=N. So g⁻¹N acts as inverse of gN. Normality ensures coset multiplication is well-defined.
N is normal in G iff gNg⁻¹ equals
A G
B {e}
C N
D gN
Explanation: Conjugation invariance means gNg⁻¹=N for every g in G. This is equivalent to left and right cosets matching and is a standard normality test.
The commutator of a,b is
A aba⁻¹b⁻¹
B abba
C a²b²
D a⁻¹b⁻¹
Explanation: The commutator measures non-commutativity. It equals e exactly when ab=ba. Commutators generate the commutator subgroup, important in forming abelian quotients.
Commutator subgroup G′ is generated by
A all inverses
B all identities
C all commutators
D all cosets
Explanation: G′ (also [G,G]) is the subgroup generated by elements aba⁻¹b⁻¹. The quotient G/G′ becomes abelian, capturing how far G is from being commutative.
A group G is abelian iff commutator subgroup is
A G
B {e}
C of index 2
D cyclic always
Explanation: If G is abelian, every commutator is e, so G′={e}. Conversely, if G′ is trivial, all commutators are e, which forces ab=ba for all a,b.
For homomorphism φ, kernel equals preimage of
A identity of G
B zero divisors
C image of φ
D identity of H
Explanation: ker(φ) is the set of g in G with φ(g)=e_H. It captures which elements become “invisible” under φ and forms a normal subgroup automatically.
For homomorphism φ, φ(e_G) equals
A 0 always
B e_H
C e_G
D undefined
Explanation: φ(e_G)=φ(e_G·e_G)=φ(e_G)φ(e_G). Multiply by inverse in H to conclude φ(e_G)=e_H. Homomorphisms always send identity to identity.
For homomorphism φ, φ(g⁻¹) equals
A φ(g)⁻¹
B φ(g)
C φ(g)²
D e_H always
Explanation: Since gg⁻¹=e_G, apply φ: φ(g)φ(g⁻¹)=φ(e_G)=e_H. Hence φ(g⁻¹) must be inverse of φ(g). This helps compute images easily.
If kerφ = G, then φ is
A injective map
B isomorphism
C automorphism
D trivial map
Explanation: If every element maps to identity, then kernel is entire G. That means φ(g)=e_H for all g, so it is the trivial homomorphism with image {e_H}.
If imφ = {e_H}, then φ is
A injective map
B surjective map
C trivial map
D bijection
Explanation: Image containing only identity means every element maps to identity, so the homomorphism is trivial. Then kernel is all of G and it cannot be injective unless G is trivial.
In ring homomorphism f: R→S, f(0_R) equals
A 0_S
B 1_S
C 0_R
D not fixed
Explanation: f(0)=f(0+0)=f(0)+f(0). Subtract f(0) from both sides using additive inverses in S to get f(0)=0. So zero maps to zero.
In unital ring homomorphism, f(1_R) equals
A 0_S
B 1_S
C −1_S
D any unit
Explanation: A unital (unity-preserving) ring homomorphism is defined to send 1_R to 1_S. This condition is common in algebra and ensures strong structure preservation.
Units in a ring are elements with
A additive inverse only
B zero divisor property
C multiplicative inverse
D idempotent property
Explanation: A unit u has some v with uv=vu=1. Units form a group under multiplication. In ℤ, only 1 and −1 are units; in ℤₙ, units are coprime classes.
In ℤ, associates are numbers differing by
A a unit factor
B a prime factor
C a gcd factor
D a square factor
Explanation: In ℤ, units are ±1. So two integers are associates if they differ by multiplication by ±1, like 5 and −5. They generate the same principal ideal.
In ℤ, gcd(a,b) is generator of ideal
A aℤ ∩ bℤ
B abℤ
C aℤ − bℤ
D aℤ + bℤ
Explanation: The set aℤ+bℤ = {ax+by} is an ideal in ℤ. It equals dℤ where d=gcd(a,b). This is Bézout’s identity in ideal form.
In ℤ, lcm(a,b) generates ideal
A aℤ + bℤ
B (a+b)ℤ
C aℤ ∩ bℤ
D (ab)ℤ only
Explanation: The intersection contains integers divisible by both a and b, i.e., multiples of lcm(a,b). Hence aℤ∩bℤ = lcm(a,b)ℤ, matching divisibility properties.
In a field, only ideals are
A all subgroups
B {0} and F
C all subsets
D only maximal ideals
Explanation: In a field, any nonzero ideal contains 1 because it contains inverse of a nonzero element and absorbs multiplication. If 1 is in the ideal, the ideal equals the whole field.
In integral domain, product ab=0 implies
A a and b units
B a=b always
C both nonzero
D a=0 or b=0
Explanation: “No zero divisors” means a nonzero times nonzero cannot give zero. So if ab=0 in an integral domain, at least one factor must be zero, like in integers.
ℤ₁₂ has zero divisors because
A 3·4 ≡0
B 5·5 ≡0
C 7·7 ≡0
D 11·11 ≡0
Explanation: In ℤ₁₂, 3 and 4 are nonzero classes, but 3×4=12≡0 (mod 12). This shows zero divisors exist, so ℤ₁₂ is not an integral domain.
ℤ₁₁ is a field because 11 is
A even
B composite
C prime
D square
Explanation: ℤ_p is a field for prime p since every nonzero class has an inverse mod p. With p=11, there are no zero divisors, and division works for all nonzero residues.
Field characteristic cannot be
A prime number
B composite number
C zero
D positive prime
Explanation: If a field had composite characteristic mn, then (m·1)(n·1)=0 would create zero divisors, impossible in a field. So characteristic is either 0 or a prime.
In a field of characteristic p, sum of 1 repeated p times is
A 0
B 1
C p
D undefined
Explanation: Characteristic p means p·1=0 by definition. This causes modular-like behavior inside the field. For example, in ℤ_p, adding 1 p times returns to 0.
A subfield must contain
A only 0
B only units
C only primes
D 1 and 0
Explanation: A subfield is a subset that is itself a field with the same operations. So it must contain additive identity 0 and multiplicative identity 1 and be closed under inverses.
In field F, equation ax=b with a≠0 has
A two solutions
B no solution
C unique solution
D infinite solutions
Explanation: Multiply both sides by a⁻¹ to get x=a⁻¹b. Existence of multiplicative inverses for all nonzero elements is the key field property ensuring unique solvability.
In ring of 2×2 matrices over ℝ, multiplication is
A noncommutative
B commutative
C idempotent always
D always zero
Explanation: Matrix multiplication is associative and distributive, but generally AB≠BA. Therefore the ring of matrices is typically noncommutative, though it has identity matrix as unity.
Identity element in 2×2 matrix ring is
A zero matrix
B identity matrix
C any diagonal matrix
D any invertible matrix
Explanation: The identity matrix I satisfies IA=AI=A for every 2×2 matrix A. So it acts as the multiplicative identity (unity) in the matrix ring.
Ideal in ℤₙ correspond to divisors of
A φ(n)
B n−1
C n
D n+1
Explanation: Every ideal in ℤₙ is generated by a divisor d of n and has form ⟨d⟩. Different divisors give different ideals, linking ideal structure directly to divisibility of n.
In ℤₙ, ideal generated by d consists of all
A numbers coprime to n
B prime residues only
C nonzero units
D multiples of d
Explanation: The ideal ⟨d⟩ in ℤₙ contains all classes of the form d·k (mod n). It is closed under addition and absorbs multiplication by any residue, matching ideal definition.
Quotient ring (ℤ)/(nℤ) is isomorphic to
A ℤₙ
B ℚ
C ℝ
D nℤ
Explanation: Modding integers by nℤ identifies numbers differing by multiples of n. The resulting classes are exactly residues mod n. Hence ℤ/nℤ ≅ ℤₙ as rings.
A homomorphism preserves identity in groups means
A φ(e)=0
B φ(e)=e
C φ(e)=g
D not necessary
Explanation: Group homomorphisms always send identity to identity because φ(e)=φ(e·e)=φ(e)φ(e). Canceling gives φ(e)=e. This is a basic property used often.
If G is cyclic, every subgroup of G is
A normal only
B prime order
C cyclic
D trivial always
Explanation: Any subgroup of a cyclic group is generated by some power of the generator. In finite cyclic groups, there is exactly one subgroup for each divisor of the group order.
If N is normal in G, then G/N is abelian when N contains
A commutator subgroup
B center of G
C trivial subgroup
D generator only
Explanation: The quotient G/N is abelian exactly when all commutators become identity in the quotient. That happens when the commutator subgroup G′ is contained in N.