If |G|=pq with p<q primes and p ∤ (q−1), then G is
A simple
B nonabelian always
C no subgroups
D cyclic
By Sylow results, the Sylow-q subgroup is normal and unique, and the Sylow-p subgroup is also normal when p does not divide (q−1). Then G becomes direct product of cyclic groups, hence cyclic.
If H has index p (smallest prime dividing |G|), then H is
A normal
B trivial
C cyclic only
D maximal ideal
The action of G on left cosets gives a homomorphism into S_p. Kernel lies in H. With p minimal prime divisor, the image size forces kernel to be all of H, making H normal.
Number of elements of order d in cyclic group of order n equals
A φ(n) always
B φ(d) if d|n
C d always
D n−d always
In a cyclic group, each divisor d of n corresponds to a unique subgroup of size d. That subgroup has exactly φ(d) generators, which are precisely the elements of order d.
In cyclic group of order 18, number of elements of order 9 is
A 9
B 3
C 6
D 2
Elements of order 9 exist since 9 divides 18. The number of elements of order 9 equals φ(9)=9(1−1/3)=6. This uses the cyclic group counting rule.
In cyclic group of order 20, number of elements of order 10 is
A 4
B 10
C 2
D 8
Since 10 divides 20, elements of order 10 exist. Count is φ(10)=10(1−1/2)(1−1/5)=4. They are generators of the unique subgroup of order 10.
If a and b commute and ord(a)=m, ord(b)=n with gcd(m,n)=1, then ord(ab) is
A m+n
B max(m,n)
C gcd(m,n)
D mn
When a and b commute, (ab)^k=a^k b^k. If gcd(m,n)=1, the smallest k making both a^k=e and b^k=e is k=mn, so ord(ab)=mn.
If N ◁ G, then quotient map π:G→G/N is
A injective homomorphism
B not a homomorphism
C surjective homomorphism
D always isomorphism
π(g)=gN covers every coset, so it is onto. Also π(gh)=(gh)N=(gN)(hN), so it preserves operation. It is injective only when N is trivial.
If φ:G→H is homomorphism and imφ has size 1, then kerφ is
A G
B {e}
C trivial subgroup
D center of G
If image is only identity, then φ(g)=e for all g, so every element maps to identity. Hence kernel is entire G. This is exactly the trivial homomorphism case.
If kerφ ⊆ Z(G), then φ factors through which quotient
A G/Z(G)
B H/imφ
C G/kerφ
D G/G′
Any homomorphism naturally induces an injective map from G/kerφ to imφ. Containment in the center is extra information but the standard factorization always goes through G/kerφ.
If N contains commutator subgroup G′, then G/N is
A cyclic
B simple
C trivial always
D abelian
In G/N, commutators become N, i.e., identity coset. So (gN)(hN)=(hN)(gN) for all g,h. Thus G/N is abelian exactly when G′ ⊆ N.
The smallest normal subgroup containing a set S is called
A centralizer
B index set
C normal closure
D kernel image
Normal closure of S is the subgroup generated by all conjugates gsg⁻¹ (s in S, g in G). It is the smallest normal subgroup of G containing S.
In any group, conjugacy class size of x equals index of
A center Z(G)
B centralizer C(x)
C commutator subgroup
D kernel of π
The orbit-stabilizer idea gives |Cl(x)| = [G : C(x)], where C(x) are elements commuting with x. This links conjugation action to subgroup indices.
If x is in center Z(G), then its conjugacy class size is
A 1
B |G|
C index of Z(G)
D prime always
If x commutes with all elements, then gxg⁻¹=x for every g. So the conjugacy class contains only x. Central elements have singleton conjugacy classes.
In a finite p-group, center Z(G) is
A always trivial
B never a subgroup
C nontrivial
D equal to G always
In a finite p-group, class equation shows |Z(G)| is divisible by p, so at least p elements. Hence Z(G) cannot be {e}. This is key in p-group theory.
If |G|=p² (p prime), then G must be
A simple
B nonabelian always
C with trivial center
D abelian
For order p², the center is nontrivial. If center has order p², G is abelian. If center has order p, quotient G/Z(G) has order p, hence cyclic, forcing G abelian.
If G/Z(G) is cyclic, then G is
A abelian
B simple
C cyclic always
D of prime order
If G/Z(G)=⟨gZ(G)⟩, then every element is g^k z. Using centrality of z, products commute. This standard result shows cyclic quotient by center forces abelian group.
If H ◁ G and K ◁ G, then H∩K is
A only subgroup
B not closed
C normal in G
D always trivial
For x in H∩K and g in G, gxg⁻¹ lies in H since H normal and in K since K normal. Thus conjugation keeps it in intersection, so H∩K is normal.
If H ◁ G and K is subgroup, then HK is subgroup when
A |H|=|K|
B G is cyclic
C K is normal
D HK = KH
HK is a subgroup if it is closed under multiplication and inverses. A sufficient condition is HK=KH, which ensures closure. If K is normal, then HK=KH automatically.
If H ◁ G and K ◁ G, then HK is
A never subgroup
B normal in G
C only a coset
D always cyclic
When both are normal, hk conjugated by g gives (ghg⁻¹)(gkg⁻¹) still in HK. Also HK is a subgroup. Hence HK is a normal subgroup of G.
Second isomorphism theorem requires K to be
A cyclic only
B of prime order
C normal in G
D equal to H
The theorem states HK/K ≅ H/(H∩K) provided K is normal (at least in HK, commonly assumed normal in G). Normality makes coset quotient and mapping well-defined.
For normal N, quotient G/N is simple iff N is
A maximal normal
B minimal normal
C cyclic subgroup
D trivial subgroup
A normal subgroup corresponds to a normal subgroup in the quotient. G/N has no nontrivial normal subgroups exactly when there is no normal subgroup strictly between N and G, i.e., N is maximal normal.
A group homomorphism with trivial kernel is called
A epimorphism
B endomorphism
C monomorphism
D trivial map
Trivial kernel implies injective, and an injective homomorphism is a monomorphism. It embeds G into H without collapsing distinct elements.
A surjective homomorphism is called
A monomorphism
B epimorphism
C automorphism
D endomorphism
Epimorphism means onto. It maps the domain onto the whole codomain. If it is also injective, it becomes an isomorphism.
If φ:G→H is onto and |kerφ|=k, then |G| equals
A |H|/k
B |H|+k
C k−|H|
D k|H|
By first isomorphism theorem, |G/kerφ|=|H|. So |G|=|kerφ|·|H| for finite groups. This is a counting form of the theorem.
In ring theory, kernel of ring homomorphism is always
A a subgroup only
B a coset
C an ideal
D a field
Kernel is closed under subtraction and absorbs multiplication by any ring element from both sides (in commutative case). Hence it satisfies the ideal conditions, enabling quotient rings and isomorphism theorems.
First isomorphism theorem for rings says
A R/kerf ≅ imf
B kerf ≅ imf
C R ≅ kerf
D imf ≅ S/kerf
The natural map from R to imf has kernel kerf. Factoring R by the kernel removes exactly the collapsed part and produces a ring isomorphic to the image, preserving addition and multiplication.
Ideal ⟨a,b⟩ in a commutative ring consists of
A ab only
B a+b only
C a−b only
D ra+sb
The ideal generated by a and b is the set of all finite combinations ra+sb with r,s in R. It is the smallest ideal containing both a and b.
In ℤ, ideal ⟨6,15⟩ equals
A 1ℤ
B 30ℤ
C 3ℤ
D 9ℤ
In ℤ, ⟨a,b⟩ = gcd(a,b)ℤ. gcd(6,15)=3, and combinations 6x+15y produce exactly multiples of 3. So the ideal equals 3ℤ.
In ℤ, ideal ⟨4,10⟩ equals
A 2ℤ
B 4ℤ
C 14ℤ
D 40ℤ
gcd(4,10)=2, and Bézout gives integers x,y with 4x+10y=2. So ⟨4,10⟩ contains 2 and hence all multiples of 2, exactly 2ℤ.
In ℤₙ, ideal generated by class [d] has size
A gcd(n,d)
B n/gcd(n,d)
C φ(n)
D n−d
In ℤₙ, ⟨[d]⟩ corresponds to multiples of d mod n and equals ⟨g⟩ where g=gcd(n,d). The ideal has g distinct elements: 0,g,2g,…,(g−1)g (mod n).
In ℤ₁₂, size of ideal ⟨[4]⟩ is
A 4
B 6
C 3
D 12
Here n=12 and d=4, gcd(12,4)=4. So |⟨[4]⟩|=4. Elements are {0,4,8, (12)≡0} giving exactly four distinct classes: 0,4,8,? actually multiples: 0,4,8.
In ℤ₁₂, ideal ⟨[4]⟩ equals
A {0,4,8}
B {0,6}
C {1,5,7,11}
D all classes
Multiples of 4 mod 12 are 0,4,8, then repeat. So the ideal is {0,4,8}. It is closed under addition and absorbs multiplication by any class in ℤ₁₂.
In ℤ₁₂, size of ideal ⟨[4]⟩ is actually
A 4
B 6
C 12
D 3
The ideal ⟨[d]⟩ has n/gcd(n,d) distinct multiples of d mod n. For n=12, d=4, gcd=4, so size is 12/4=3, matching {0,4,8}. (Earlier shortcut can confuse size.)
In a PID, every ideal is
A maximal
B prime
C principal
D zero ideal
A principal ideal domain is defined by the property that every ideal is generated by a single element. This simplifies ideal theory and supports gcd-like arguments.
In an integral domain, irreducible implies prime when domain is
A any domain
B UFD
C Boolean ring
D noncommutative ring
In a unique factorization domain, irreducible elements behave like primes: if an irreducible divides a product, it divides one factor. This relies on uniqueness of factorization.
In a field F, ideal ⟨f(x)⟩ in F[x] is maximal iff f(x) is
A constant polynomial
B divisible by x
C zero polynomial
D irreducible
In F[x], a principal ideal ⟨f⟩ is maximal exactly when the quotient F[x]/⟨f⟩ is a field, which holds precisely when f is irreducible over F.
Field extension F[x]/⟨f⟩ has degree equal to
A deg f
B φ(deg f)
C |F|
D gcd coefficients
When f is irreducible over field F, the quotient F[x]/⟨f⟩ is a field where classes of polynomials reduce to degree < deg f. This gives an F-vector space basis of size deg f.
A finite field with p² elements can be constructed as
A ℤ_{p²} always
B ℚ[x]/⟨x²⟩
C ℤ_p[x]/⟨irreducible deg2⟩
D ℤ_p×ℤ_p only
To get p² elements, take polynomials mod an irreducible quadratic over ℤ_p. The quotient becomes a field with p² elements. ℤ_{p²} is not a field for p>1.
Field of order 4 is isomorphic to
A ℤ₄
B ℤ₂[x]/⟨x²+x+1⟩
C ℤ₂×ℤ₂ as field
D ℚ/4ℚ
Over ℤ₂, x²+x+1 has no roots, hence irreducible. Quotient by this ideal gives a field with 2²=4 elements. ℤ₄ has zero divisors, so not a field.
In a field, any nonzero homomorphism f:F→K is
A injective
B never injective
C always zero map
D not defined
Kernel of a field homomorphism is an ideal of F. Field has only {0} and F as ideals. Nonzero homomorphism cannot have kernel F, so kernel is {0}, hence injective.
A ring homomorphism maps unity 1 to 1 only if it is
A trivial
B surjective
C injective
D unital
Some ring homomorphisms are defined to preserve multiplicative identity; these are called unital homomorphisms. Without this assumption, a homomorphism may send 1 to an idempotent element.
In S₄, number of 3-cycles is
A 6
B 12
C 8
D 4
A 3-cycle chooses 3 elements out of 4: C(4,3)=4 ways. For each chosen triple, there are 2 distinct 3-cycles. Total 4×2=8.
In S₄, order of a permutation (12)(34) is
A 2
B 4
C 6
D 8
(12)(34) is product of two disjoint transpositions. Each transposition has order 2, and disjoint cycles’ order is lcm(2,2)=2. Applying twice returns identity.
In S₄, order of a 4-cycle is
A 2
B 4
C 6
D 8
A 4-cycle rotates four elements in a loop. It returns to identity after 4 applications and not earlier, so its order equals the cycle length 4.
In ℤₙ, the group of units (ℤₙ)× is cyclic for n=
A 15
B 16
C 14
D 12
(ℤₙ)× is cyclic for n=1,2,4,p^k,2p^k where p is odd prime. Here 16=2^4 fits p^k form? actually 2^k gives cyclic only for k≤2; for 16 it is not cyclic.
(ℤ₁₆)× is cyclic?
A Yes
B Only if prime
C Only if odd
D No
The unit group modulo 2^k is cyclic only for k=1 or 2 (mod 2 or 4). For 16 (k=4), (ℤ₁₆)× is not cyclic; it has structure like C₂×C₂×C₂.
The correct classification for cyclic (ℤₙ)× includes n=
A 2p^k form
B 6 always
C n odd only
D n composite only
The unit group (ℤₙ)× is cyclic exactly when n is 1,2,4,p^k, or 2p^k for odd prime p. This is a standard theorem in number theory.
If a has order 15, then a^10 has order
A 5
B 3
C 15
D 1
ord(a^k)=n/gcd(n,k). Here n=15, k=10, gcd(15,10)=5, so ord(a^10)=15/5=3. So correct is 3, not 5.
If a has order 15, then order of a^10 is
A 5
B 15
C 3
D 1
Using ord(a^k)=n/gcd(n,k) with n=15 and k=10 gives gcd=5, so order is 15/5=3. This checks by (a^10)^3=a^30=(a^15)^2=e.
In a group, number of solutions to x^2=e equals number of elements of
A order 1 or 2
B order 3
C prime order
D infinite order
x^2=e means x has order dividing 2. So solutions are exactly identity (order 1) plus all elements of order 2. Counting solutions equals counting involutions plus identity.