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A group homomorphism h between groups G and H must satisfy h(u * v) = h(u) + h(v) for all u, v in G.
Answer: False
The defining property of a group homomorphism h: G โ H is that it preserves the group operation, meaning h(u * v) = h(u) ยท h(v) for all u, v in G, where '*' is the operation in G and 'ยท' is the operation in H. The use of '+' for the operation in H implies a specific case (an additive group), not a universal requirement.
A group homomorphism h: G -> H always maps the identity element of G to the identity element of H.
Answer: True
This is a fundamental property. For any homomorphism h: G โ H, if e_G is the identity in G and e_H is the identity in H, then h(e_G) = e_H. This can be proven by considering h(u) = h(u * e_G) = h(u) ยท h(e_G) and multiplying by the inverse of h(u).
If h is a group homomorphism, then h(uโปยน) is equal to h(u)โปยน for any element u in the domain group.
Answer: True
This property follows from the definition of a homomorphism and group inverses. Since h(u * uโปยน) = h(e_G) = e_H and h(u * uโปยน) = h(u) ยท h(uโปยน), it implies h(uโปยน) = (h(u))^{-1}.
A homomorphism preserves the group operation, meaning h(u) * h(v) = h(u * v).
Answer: True
This statement correctly captures the defining characteristic of a group homomorphism: it respects the group operation, ensuring h(u * v) = h(u) ยท h(v).
In the context of group homomorphisms, what does the term 'compatible with the group structure' imply?
Answer: False
While 'compatible with the group structure' primarily means preserving the group operation (h(u*v) = h(u)h(v)), it also implicitly encompasses other essential properties like mapping the identity element to the identity and inverses to inverses. Stating it implies *only* operation preservation is incomplete.
What is the fundamental property that defines a group homomorphism h: G -> H?
Answer: h(u * v) = h(u) ยท h(v) for all u, v in G.
The defining characteristic of a group homomorphism is its ability to preserve the group operation. This means that for any elements u and v in the domain group G, the image of their product under the homomorphism is equal to the product of their images in the codomain group H.
According to the definition, what must a group homomorphism h do with the identity elements of the groups G and H?
Answer: Map e_G to e_H.
A group homomorphism must map the identity element of the domain group (e_G) to the identity element of the codomain group (e_H). This property, h(e_G) = e_H, is a direct consequence of the homomorphism property.
Which property must a group homomorphism h satisfy regarding element inverses?
Answer: h(uโปยน) = (h(u))โปยน.
A group homomorphism preserves the inverse property. For any element u in the domain group G, the image of its inverse, h(uโปยน), must be equal to the inverse of the image of u, (h(u))โปยน.
What does it mean for a function to be 'compatible with the group structure'?
Answer: It respects the group operation, i.e., h(u*v) = h(u)ยทh(v).
A function is considered compatible with the group structure if it preserves the group operation. For a homomorphism h: G โ H, this means h(u*v) = h(u) ยท h(v) for all u, v โ G.
The kernel of a group homomorphism h: G -> H is the set of elements in H mapped to the identity element of G.
Answer: False
The kernel of a group homomorphism h: G โ H, denoted ker(h), is the set of elements in the domain group G that map to the identity element of the codomain group H (i.e., {u โ G | h(u) = e_H}).
The kernel of any group homomorphism is always a normal subgroup of the domain group.
Answer: True
This is a fundamental property: for any homomorphism h: G โ H, the kernel ker(h) is a normal subgroup of the domain group G.
The image of a group homomorphism h: G -> H is always a subgroup of the domain group G.
Answer: False
The image of a group homomorphism h: G โ H, denoted im(h), is always a subgroup of the codomain group H, not necessarily the domain group G.
A group homomorphism is injective if and only if its kernel contains only non-identity elements.
Answer: False
A group homomorphism h is injective if and only if its kernel is the trivial subgroup {e_G}. If the kernel contains any non-identity elements, those elements map to the identity, violating injectivity.
The kernel of a group homomorphism h: G -> H is formally defined as:
Answer: {u โ G | h(u) = e_H}
The kernel of a homomorphism h: G โ H is the set of all elements in the domain G that map to the identity element e_H in the codomain H.
What is the relationship between the kernel of a homomorphism and injectivity?
Answer: h is injective if ker(h) = {e_G}.
A group homomorphism h is injective if and only if its kernel consists solely of the identity element of the domain group. If ker(h) = {e_G}, then h(u) = h(v) implies u = v.
The image of a group homomorphism h: G -> H is always:
Answer: A subgroup of the codomain H.
The image of a homomorphism, defined as the set of all outputs {h(u) | u โ G}, forms a subgroup of the codomain H. This is because it is closed under the group operation, contains the identity element of H, and is closed under inverses.
The kernel of a group homomorphism h: G -> H is always a:
Answer: Normal subgroup of G.
The kernel of any group homomorphism is guaranteed to be a normal subgroup of the domain group G. This property is fundamental in establishing quotient groups and the First Isomorphism Theorem.
If h: G -> H is a group homomorphism, which statement about ker(h) and im(h) is correct?
Answer: ker(h) is a normal subgroup of G, and im(h) is a subgroup of H.
The kernel of a homomorphism is always a normal subgroup of the domain, while the image is always a subgroup of the codomain. These are fundamental properties derived from the definition of a homomorphism.
A monomorphism is a group homomorphism that is surjective but not necessarily injective.
Answer: False
A monomorphism is defined as an injective (one-to-one) group homomorphism. Surjectivity is not a required condition for a monomorphism.
An epimorphism is a group homomorphism that maps distinct elements to distinct elements.
Answer: False
An epimorphism is a group homomorphism that is surjective (onto). The property of mapping distinct elements to distinct elements defines injectivity, which characterizes monomorphisms.
A group isomorphism is a homomorphism that is both injective and surjective.
Answer: True
This is the correct definition of a group isomorphism: a homomorphism that is bijective, meaning it is both injective and surjective. This establishes a structural equivalence between the groups.
Isomorphic groups are structurally identical.
Answer: True
The term 'isomorphic' signifies structural identity. Isomorphic groups share all algebraic properties and are equivalent from an abstract algebraic viewpoint.
A group homomorphism that is one-to-one (injective) is called a:
Answer: Monomorphism
A homomorphism that is injective is known as a monomorphism. This property ensures that distinct elements in the domain map to distinct elements in the codomain.
What defines an epimorphism in group theory?
Answer: A homomorphism that is surjective.
An epimorphism is a group homomorphism that is surjective, meaning that every element in the codomain group is the image of at least one element from the domain group.
Which condition must be met for a group homomorphism h to be an isomorphism?
Answer: It must be bijective (both injective and surjective).
An isomorphism is a homomorphism that is bijective, meaning it is both injective (one-to-one) and surjective (onto). This establishes a one-to-one correspondence between the elements of the two groups while preserving their structure.
Which of the following is NOT necessarily true for *all* group homomorphisms?
Answer: h is injective.
While homomorphisms preserve the group operation, map identity to identity, and map inverses to inverses, injectivity is not a guaranteed property. Injectivity is specific to monomorphisms and isomorphisms.
Which type of homomorphism is both injective and surjective?
Answer: Isomorphism
A homomorphism that is both injective (one-to-one) and surjective (onto) is defined as an isomorphism. Isomorphisms indicate that two groups are structurally identical.
In the context of group theory, what does the term 'structurally identical' imply?
Answer: The groups are isomorphic.
When two groups are described as structurally identical, it signifies that there exists an isomorphism between them. This means they possess the same algebraic properties and structure, differing only in the labels of their elements.
An endomorphism is a homomorphism from a group G to a different group H.
Answer: False
An endomorphism is specifically a homomorphism where the domain and codomain are the same group (G โ G). A homomorphism between different groups G and H is simply termed a homomorphism.
An automorphism is a bijective homomorphism from a group onto itself.
Answer: True
This is the correct definition of an automorphism: a homomorphism that is bijective and maps a group to itself, preserving its structure.
The set of all automorphisms of a group G forms a group under the operation of functional composition.
Answer: True
The set of automorphisms of a group G, denoted Aut(G), forms a group under the operation of function composition. This group is known as the automorphism group of G.
The set of homomorphisms between two abelian groups G and H forms a group under the operation of functional composition, provided G is abelian.
Answer: False
The set of homomorphisms between two abelian groups G and H, Hom(G, H), forms an abelian group under *pointwise addition*, not functional composition. Composition is relevant for the endomorphism ring.
The commutativity of the codomain group H is necessary for the sum of two homomorphisms h + k to also be a homomorphism.
Answer: True
The commutativity (abelian property) of the codomain group H is essential for the pointwise sum of two homomorphisms (h+k)(u) = h(u) + k(u) to satisfy the homomorphism property, as it allows for necessary rearrangements in the proof.
The endomorphism ring of an abelian group G uses functional composition as its addition operation.
Answer: False
The endomorphism ring of an abelian group G uses pointwise addition and functional composition. Addition is pointwise, while multiplication is composition.
The endomorphism ring of m copies of Z/nZ is isomorphic to the ring of m x m matrices over Z/nZ.
Answer: True
The endomorphism ring of the direct sum of m copies of Z/nZ, denoted End( (Z/nZ)^m ), is isomorphic to the ring of m ร m matrices with entries from Z/nZ, M_m(Z/nZ).
The automorphism group of the cyclic group of integers (Z, +) is isomorphic to the cyclic group of order 2, Z/2Z.
Answer: True
The automorphisms of (Z, +) are the identity and the negation map. This group, Aut(Z), is isomorphic to Z/2Z, the cyclic group of order 2.
A homomorphism h: G -> G is known as an:
Answer: Endomorphism
A homomorphism whose domain and codomain are the same group is called an endomorphism. It is a structure-preserving map from a group to itself.
What is a special type of endomorphism that is also an isomorphism?
Answer: Automorphism
An endomorphism that is also an isomorphism is called an automorphism. It represents a structure-preserving rearrangement of the elements within a group.
If G and H are abelian groups, the set Hom(G, H) forms:
Answer: An abelian group under pointwise addition.
The set of homomorphisms between two abelian groups G and H, denoted Hom(G, H), forms an abelian group under the operation of pointwise addition, where (h+k)(u) = h(u) + k(u).
For the sum of two homomorphisms h + k to be a homomorphism, what property must the codomain group H possess?
Answer: It must be abelian (commutative).
The commutativity of the operation in the codomain group H is necessary to prove that the sum of two homomorphisms is also a homomorphism, as it allows for required rearrangements in the proof.
The endomorphism ring of an abelian group G uses which operations?
Answer: Pointwise addition and composition.
The endomorphism ring of an abelian group G is endowed with pointwise addition and functional composition as its operations.
What is the automorphism group of the cyclic group of integers (Z, +) isomorphic to?
Answer: The cyclic group Z/2Z.
The group of automorphisms of the infinite cyclic group (Z, +), denoted Aut(Z), consists of the identity map and the map x โ -x. This group is isomorphic to Z/2Z, the cyclic group of order 2.
The first isomorphism theorem states that the image of a homomorphism is isomorphic to the quotient group G/ker(h).
Answer: True
The First Isomorphism Theorem for groups establishes that for any homomorphism h: G โ H, the quotient group G/ker(h) is isomorphic to the image of h, i.e., G/ker(h) โ im(h).
The exponential map exp(x) = e^x is a homomorphism from the multiplicative group of non-zero real numbers (R*, *) to the additive group of real numbers (R, +).
Answer: False
The exponential map exp(x) = e^x is a homomorphism from the *additive* group (R, +) to the *multiplicative* group (R*, ยท), as exp(x+y) = exp(x)exp(y). The direction and operations are reversed in the question's statement.
The kernel of the exponential map exp(z) from (C, +) to (C*, *) is the set {2ฯki | k โ Z}.
Answer: True
The exponential map exp(z) = e^z is a homomorphism from (C, +) to (C*, ยท). Its kernel consists of complex numbers z such that e^z = 1, which occurs precisely when z is an integer multiple of 2ฯi.
The kernel of ฮฆ(x) = โ2 * x from (Z, +) to (R, +) is {0}.
Answer: True
The kernel consists of integers x such that โ2 * x = 0 in (R, +). Since โ2 โ 0, the only integer solution is x = 0. Thus, the kernel is {0}.
The image of ฮฆ(x) = โ2 * x from (Z, +) to (R, +) is the set of all real numbers.
Answer: False
The image of this homomorphism is the set of all real numbers that are integer multiples of โ2, i.e., {kโ2 | k โ Z}, which is a proper subset of the real numbers.
The exponential map exp(x) = e^x from (R, +) to (R*, *) is surjective onto (R*, *).
Answer: False
The exponential map exp(x) = e^x maps from (R, +) to (R*, ยท). Its image is the set of all *positive* real numbers, not the entire set of non-zero real numbers, as it does not produce negative outputs.
According to the First Isomorphism Theorem, the image of a homomorphism is isomorphic to:
Answer: The quotient group G / ker(h).
The First Isomorphism Theorem states that for a homomorphism h: G โ H, the image of h is isomorphic to the quotient group of G by its kernel, G/ker(h). This theorem provides a profound link between homomorphisms, kernels, images, and quotient structures.
Consider the homomorphism h(u) = u mod 3 from (Z, +) to (Zโ, +). What is its kernel?
Answer: The set of all multiples of 3.
The kernel consists of integers u such that h(u) = 0 in Zโ. This means u mod 3 = 0, which implies u must be a multiple of 3. Thus, ker(h) = {..., -6, -3, 0, 3, 6, ...}.
What is the image of the homomorphism h(u) = u mod 3 from (Z, +) to (Zโ, +)?
Answer: The entire group Zโ.
The image consists of all possible outputs of h(u) = u mod 3. Since any integer modulo 3 results in 0, 1, or 2, the image is precisely the set {0, 1, 2}, which constitutes the entire group Zโ.
The exponential map exp(x) = e^x maps (R, +) to (R*, *). What is its image?
Answer: All positive real numbers.
For any real number x, e^x is always a positive real number. Conversely, for any positive real number y, there exists a real number x = ln(y) such that e^x = y. Thus, the image is the set of all positive real numbers.
What is the kernel of the exponential map exp(z) from (C, +) to (C*, *)?
Answer: {2ฯki | k โ Z}
The kernel consists of complex numbers z such that e^z = 1. This occurs precisely when z is an integer multiple of 2ฯi, i.e., z = 2ฯki for some integer k.
The homomorphism ฮฆ(x) = โ2 * x from (Z, +) to (R, +) maps integers to real numbers. What is its image?
Answer: The set {kโ2 | k โ Z}.
The image of this homomorphism consists of all real numbers that can be expressed as an integer multiple of โ2. This set is {kโ2 | k โ Z}.
Group homomorphisms serve as the morphisms in the category of groups.
Answer: True
In category theory, group homomorphisms are the structure-preserving maps (morphisms) between groups (objects). This defines the category of groups, often denoted Grp.
The category of abelian groups is an example of a non-abelian category.
Answer: False
The category of abelian groups (Ab) is a fundamental example of an *abelian category*, possessing richer structure than a general category.
The composition of two group homomorphisms is never a group homomorphism.
Answer: False
The composition of two group homomorphisms is always a group homomorphism. This property is fundamental in abstract algebra and category theory.
The category of abelian groups is classified as a(n):
Answer: Abelian category
The category of abelian groups is a foundational example of an abelian category, characterized by properties such as the existence of kernels, cokernels, and direct sums.
The composition of two group homomorphisms, g followed by f, is:
Answer: Always a homomorphism.
The composition of two structure-preserving maps (homomorphisms) is itself a structure-preserving map. This property is fundamental in abstract algebra and category theory.