2. Definition/Lemma: If φ: G 1 → G 2 is a homomorphism, the collection of elements of G 1 which φ sends to the identity of G 2 is a subgroup of G 1; it is called the kernel of φ. Thus Ker φ is certainly non-empty. you calculate the real and imaginary parts of f(x+ y) and of f(x)f(y), then equality of the real parts is the addition formula for cosine and equality of the imaginary parts is the addition formula for sine. If His a subgroup of a group Gand i: H!Gis the inclusion, then i is a homomorphism, which is essentially the statement that the group operations for H are induced by those for G. Note that iis always injective, but it is surjective ()H= G. 3. Then there exists an r2Rsuch that ˚(r) = sor equivalently that ’(r+ ker˚) = s. Thus s2im’and so ’is surjective. Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. . These are the kind of straightforward proofs you MUST practice doing to do well on quizzes and exams. Exercise Problems and Solutions in Group Theory. Note that φ(e) = f. by (8.2). Let us prove that ’is bijective. (4) For each homomorphism in A, decide whether or not it is injective. Thus ker’is trivial and so by Exercise 9, ’ is injective. Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). , φ(vn)} is a basis of W. C) For any two finite-dimensional vector spaces V and W over field F, there exists a linear transformation φ : V → W such that dim(ker(φ… Therefore a2ker˙˚. If there exists a ring homomorphism f : R → S then the characteristic of S divides the characteristic of R. Indeed, ker˚/Gso for every element g2ker˙˚ G, gker˚g 1 ˆ ker˚. . K). If (S,φ) and (S0,φ0) are two R-algebras then a ring homomorphism f : S → S0 is called a homomorphism of R-algebras if f(1 S) = 1 S0 and f φ= φ0. If r+ ker˚2ker’, then ’(r+ I) = ˚(r) = 0 and so r2ker˚or equivalently r+ ker˚= ker˚. Let s2im˚. We show that for a given homomorphism of groups, the quotient by the kernel induces an injective homomorphism. Then Ker φ is a subgroup of G. Proof. The kernel of φ, denoted Ker φ, is the inverse image of the identity. (The values of f… Decide also whether or not the map is an isomorphism. The kernel of f, defined as ker(f) = {a in R : f(a) = 0 S}, is an ideal in R. Every ideal in a ring R arises from some ring homomorphism in this way. Proof: Suppose a and b are elements of G 1 in the kernel of φ, in other words, φ(a) = φ(b) = e 2, where e 2 is the identity element of G 2.Then … Therefore the equations (2.2) tell us that f is a homomorphism from R to C . Given r ∈ R, let f be the constant function with value r. Then φ(f) = r. Hence φ is surjective. Moreover, if ˚and ˙are onto and Gis finite, then from the first isomorphism the- If a2ker˚, then ˙˚(a) = ˙(e H) = e K where e H (resp. φ is injective and surjective if and only if {φ(v1), . Solution for (a) Prove that the kernel ker(f) of a linear transformation f : V → W is a subspace of V . e K) is the identity of H (resp. This implies that ker˚ ker˙˚. Suppose that φ(f) = 0. Solution: Define a map φ: F −→ R by sending f ∈ F to its value at x, f(x) ∈ R. It is easy to check that φ is a ring homomorphism. Prove that I is a prime ideal iff R is a domain. We have to show that the kernel is non-empty and closed under products and inverses. Furthermore, ker˚/ker˙˚. functions in F vanishing at x. (b) Prove that f is injective or one to one if and only… The homomorphism f is injective if and only if ker(f) = {0 R}. The function f: G!Hde ned by f(g) = 1 for all g2Gis a homo-morphism (the trivial homomorphism). 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