Explanation: The inverse of a⁻¹ is the element that multiplies with a⁻¹ to give identity. Since a⁻¹·a=e and a·a⁻¹=e, the inverse of a⁻¹ must be a.
If a has order 9, then a⁶ has order
A 3
B 9
C 6
D 1
Explanation: For finite order n, ord(a^k)=n/gcd(n,k). Here n=9 and k=6, gcd(9,6)=3, so ord(a⁶)=9/3=3.
If ord(a)=8, then ord(a⁶) is
A 8
B 4
C 2
D 1
Explanation: Use ord(a^k)=n/gcd(n,k). With n=8 and k=6, gcd(8,6)=2, so ord(a⁶)=8/2=4. This relies on properties of cyclic subgroups.
In a group, equation a⁻¹b=a⁻¹c implies
A a=b
B b=a
C b=c
D c=e
Explanation: Multiply both sides on the left by a: a(a⁻¹b)=b and a(a⁻¹c)=c. Since equality is preserved, b=c. This is left cancellation using inverses.
In a group, if ab=e then b equals
A a⁻¹
B a
C e
D b⁻¹
Explanation: If ab=e, multiply by a⁻¹ on the left to get b=a⁻¹e=a⁻¹. This shows the inverse is unique and also that b must be inverse of a.
In a group, if ab=ac and a≠e, then
A b=e
B c=e
C ab=e
D b=c
Explanation: Left cancellation works for any a, not only a≠e. Multiply by a⁻¹ on the left: a⁻¹ab=b and a⁻¹ac=c, so b=c. The condition a≠e is unnecessary.
If H and K are subgroups of abelian group G, then HK is
A never subgroup
B always empty
C a subgroup
D only a coset
Explanation: In abelian groups, the product set HK={hk} is closed and equals KH, making it a subgroup. Commutativity helps ensure closure under inverses and products within HK.
Intersection of two subgroups H∩K is always
A a subgroup
B a coset
C not closed
D empty always
Explanation: H∩K contains identity and is closed under operation and inverses because both H and K have these properties. Thus the intersection of any subgroups is itself a subgroup.
Union H∪K of two subgroups is a subgroup when
A both finite only
B orders are coprime
C one contains other
D both cyclic only
Explanation: If neither subgroup contains the other, union fails closure: pick h in H\K and k in K\H, then hk is in neither generally. If one contains the other, union equals the larger subgroup.
If |G|=21, then any element order can be
A 4
B 5
C 8
D 7
Explanation: By Lagrange, element order must divide 21. Divisors are 1,3,7,21. Hence 7 is possible. Orders like 4,5,8 cannot occur in any group of order 21.
If G has order 15, subgroup orders possible include
A 6
B 10
C 5
D 14
Explanation: Subgroup order must divide 15, so possible orders are 1,3,5,15. Therefore a subgroup of order 5 is possible, while 6,10,14 do not divide 15.
In a cyclic group of order 12, number of elements of order 6 is
A 6
B 2
C 4
D 1
Explanation: In cyclic group of order n, number of elements of order d is φ(d) if d divides n. Here d=6 divides 12, so count is φ(6)=2.
In cyclic group of order 12, number of elements of order 4 is
A 2
B 4
C 6
D 1
Explanation: Elements of order 4 exist because 4 divides 12. The number is φ(4)=2. In ⟨g⟩, the elements g^3 and g^9 have order 4.
If H is subgroup of index 3 in G, then |G| equals
A |H|/3
B |H|+3
C 3|H|
D |H|−3
Explanation: For finite groups, index [G:H]=|G|/|H|. If index is 3, then |G|=3|H|. This comes from coset partition into 3 equal blocks.
Left cosets of H partition G means
A overlap always
B only two cosets
C equal to H only
D disjoint cover of G
Explanation: Every g in G lies in exactly one left coset gH, and distinct cosets do not intersect. So cosets form a partition: they cover G and are pairwise disjoint.
If N is normal, coset multiplication (gN)(hN) depends on
A only g,h cosets
B chosen representatives
C element order only
D subgroup index only
Explanation: Normality makes coset product well-defined: any other representatives g’ in gN and h’ in hN give the same product coset. Without normality, result can change.
If N is not normal, then (gN)(hN) may be
A always identity
B always commutative
C not well-defined
D always subgroup
Explanation: If N is not normal, different representatives of the same coset can produce different resulting cosets after multiplication. Hence you cannot define a consistent group operation on cosets.
A subgroup N is normal if and only if N is kernel of
A some isomorphism only
B some bijection
C some subset map
D some homomorphism
Explanation: Kernels are always normal. Conversely, given a normal subgroup N, the natural projection π:G→G/N has kernel N. So normal subgroups correspond exactly to kernels.
Natural projection map π:G→G/N is defined by
A π(g)=Ng
B π(g)=gN
C π(g)=g⁻¹N
D π(g)=N only
Explanation: Projection sends each element to its coset. It is a surjective homomorphism: π(gh)=(gh)N=(gN)(hN). Its kernel is exactly N.
Under projection π:G→G/N, kernel equals
A {e}
B G
C N
D center
Explanation: Kernel consists of elements g with π(g)=N, meaning gN=N, which happens exactly when g is in N. So ker(π)=N.
If φ:G→H is homomorphism, then image imφ is always
A a subgroup of H
B a subgroup of G
C an ideal
D a coset only
Explanation: Image is closed under operation and inverses because φ(ab)=φ(a)φ(b) and φ(a⁻¹)=φ(a)⁻¹. It contains identity, so it forms a subgroup of H.
If φ is injective, then G is isomorphic to
A kerφ
B H/kerφ
C imφ
D G/φ(G)
Explanation: If φ is one-to-one, then it gives a bijective homomorphism between G and its image. So G ≅ imφ. This is often called an isomorphism onto its image.
If φ is surjective, then H is isomorphic to
A kerφ
B G/kerφ
C imφ/kerφ
D G×kerφ
Explanation: By the first isomorphism theorem, G/kerφ ≅ imφ. If φ is onto, imφ=H, so H ≅ G/kerφ. Quotient captures exactly the “collapsed” part.
If kerφ={e} and φ is onto, then φ is
A trivial map
B endomorphism
C zero map
D isomorphism
Explanation: Kernel {e} means injective. Onto means surjective. Together they mean bijective homomorphism, i.e., an isomorphism. So the structures are the same up to renaming elements.
In groups, “kernel” concept links to normal subgroup because
A kernel is always cyclic
B kernel is never subgroup
C kernel is normal
D kernel equals image
Explanation: For any g in G and k in kerφ, φ(gkg⁻¹)=φ(g)φ(k)φ(g)⁻¹=e, so gkg⁻¹ stays in kernel. This conjugation closure makes kernel normal.
In a ring, an ideal I must be closed under
A subtraction
B division
C inversion
D square roots
Explanation: Ideals are additive subgroups, so they must be closed under addition and additive inverses, hence subtraction. This is essential so cosets a+I form a quotient ring under addition.
In ring R, if I is ideal and a in I, r in R, then ra is
A in R\I always
B always 0
C always unit
D in I
Explanation: Absorption property defines ideals: multiplying any ring element by an element of the ideal remains in the ideal. This ensures the quotient ring multiplication is well-defined.
In ℤ, ideal 6ℤ contains
A 7
B 5
C 18
D 1
Explanation: 6ℤ consists of all multiples of 6. Since 18=6×3, it is in 6ℤ. Numbers like 7,5,1 are not multiples of 6, so they are not in the ideal.
In ℤₙ, number of units equals
A φ(n)
B n
C n−1
D gcd(n,2)
Explanation: Units in ℤₙ are residue classes coprime to n. The count of integers 1≤a≤n with gcd(a,n)=1 is Euler’s totient φ(n), so |(ℤₙ)×|=φ(n).
In ℤ₁₂, number of units is
A 6
B 4
C 8
D 12
Explanation: Units mod 12 are numbers coprime to 12: 1,5,7,11. That is 4 units, matching φ(12)=12(1−1/2)(1−1/3)=4.
In ℤ₁₂, element [5] has inverse equal to
A [7]
B [11]
C [5]
D [3]
Explanation: Check 5×5=25≡1 (mod 12). So [5] is its own inverse. This happens when a²≡1 mod n, which can occur for some units.
In ℤ₁₁, inverse of [2] is
A [6]
B [5]
C [4]
D [8]
Explanation: Solve 2x≡1 (mod 11). Since 2×6=12≡1 (mod 11), [6] is the inverse. In a prime modulus field, every nonzero class has such inverse.
A ring with no zero divisors but not necessarily field is
A Boolean ring
B matrix ring
C quotient ring
D integral domain
Explanation: An integral domain is a commutative ring with unity and no zero divisors. It may lack multiplicative inverses for all nonzero elements, so it need not be a field.
In integral domain, if a|bc and gcd(a,b)=1, then
A a|b
B b|a
C a|c
D c|b
Explanation: This is Euclid’s lemma, familiar from integers and valid in many domains like UFDs. With gcd(a,b)=1, a shares no common factors with b, so divisibility passes to c.
Euclidean domain is an integral domain with
A zero divisors
B division algorithm
C no unity
D nonassociative product
Explanation: A Euclidean domain has a Euclidean function allowing division with remainder. This gives an algorithm for gcd, and implies it is a PID and a UFD, generalizing integers.
In a field, polynomial ring F[x] is
A field always
B has zero divisors
C nonassociative
D integral domain
Explanation: If F is a field, then F[x] has no zero divisors: product of nonzero polynomials is nonzero. It is not a field because nonconstant polynomials lack inverses in F[x].
Ideal ⟨x⟩ in F[x] consists of polynomials
A divisible by x
B divisible by x+1
C constant only
D irreducible only
Explanation: ⟨x⟩ is all multiples of x: x·f(x). These are exactly polynomials with zero constant term. It is a principal ideal generated by x.
In a commutative ring, quotient R/I is a field implies I is
A prime
B trivial
C maximal
D generated by 0
Explanation: The ideal I is maximal exactly when R/I is a field. If I is only prime, R/I is an integral domain, not necessarily a field. This is a key ideal characterization.
In a commutative ring, quotient R/P is integral domain implies P is
A maximal
B prime
C unit ideal
D zero ideal only
Explanation: Prime ideal P is characterized by R/P having no zero divisors. Maximal is stronger and gives a field. So “integral domain quotient” corresponds precisely to prime ideals.
In a group, center Z(G) is always
A normal subgroup
B not a subgroup
C only a coset
D always trivial
Explanation: Z(G) is a subgroup because it’s closed and contains identity. It is normal since for z in Z(G) and g in G, gzg⁻¹ still commutes with all elements.
If Z(G)=G, then G is
A cyclic
B simple
C of prime order
D abelian
Explanation: Z(G)=G means every element commutes with every element, so the group is abelian by definition. Cyclic is stronger and does not necessarily follow.
In symmetric group S₃, order of group equals
A 3
B 9
C 6
D 12
Explanation: S₃ contains all permutations of 3 objects, so it has 3! = 6 elements. It is the smallest non-abelian symmetric group and is a standard example in group theory.
In S₃, a 3-cycle has order
A 3
B 2
C 6
D 1
Explanation: A 3-cycle rotates three symbols. Applying it three times returns each symbol to its original position, and no smaller positive power does that, so its order is 3.
In S₃, a transposition has order
A 3
B 2
C 6
D 1
Explanation: A transposition swaps two elements. Doing it twice restores original arrangement, so it has order 2. This illustrates how cycle length gives order of permutation.
For permutation cycles, order equals
A sum of lengths
B product always
C lcm of lengths
D gcd of lengths
Explanation: Disjoint cycles act independently. The permutation returns to identity when each cycle completes whole rotations, so the smallest such power is the least common multiple of cycle lengths.
A group homomorphism from G onto G is called
A endomorphism
B automorphism
C monomorphism
D epimorphism
Explanation: “Onto” homomorphism is called epimorphism. Endomorphism just means map from G to G (not necessarily onto). Automorphism is bijective, and monomorphism is injective.
A group homomorphism from G into H that is one-to-one is
A monomorphism
B epimorphism
C automorphism
D trivial map
Explanation: Injective homomorphism is called monomorphism. It preserves structure without collapsing elements. If it is also onto, then it becomes an isomorphism.
In quotient group G/N, order of coset gN equals order of g in
A N only
B G modulo N
C center only
D commutator only
Explanation: The order of gN is the smallest n with (gN)^n=N, i.e., g^n in N. So it measures when powers of g land inside N, not necessarily ord(g) in G.
If N ⊆ K are normal, then factor (G/N)/(K/N) is
A isomorphic to K/N
B isomorphic to G/N
C isomorphic to G/K
D trivial always
Explanation: This is the third isomorphism theorem. Factoring by N first and then by K/N gives the same result as factoring G directly by K, simplifying chains of normal subgroups.
If a ring R has 1 and every nonzero element is unit, then R is
A field
B integral domain only
C Boolean ring
D matrix ring
Explanation: A commutative ring with unity in which every nonzero element has a multiplicative inverse is exactly a field. This ensures division is possible and eliminates zero divisors automatically