write down that formulas for all homomorphisms from z onto z12 Solution a (group
ID: 3080814 • Letter: W
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write down that formulas for all homomorphisms from z onto z12Explanation / Answer
a (group) homomorphism is a function f from one group G, to another H, usually written: f:G-->H with the following property: f(ab) = f(a)f(b) (that is, the image of a product in G, is the product of the images in H). a homomorphism is a function between groups that "respects group multiplication". most people have encountered the following homomorphism before: log:(R+,*)--->(R,+) (that is, log(xy) = log(x) + log(y)). there are a lot of useful facts about homomorphisms you probably also missed in class, and unfortunately, i don't know which ones were covered. 1. you could do this by trial-and-error, but it would be tedious. if we write Z24 = , and Z18 = , any homomorphism f would be completely determined by f(x). so we just need to figure out for which n, f(x) = y^n makes a homomorphism. one basic fact of homomorphisms is that f(x^n) = [f(x)]^n. another is that f(e) = e (convince yourself these have to be true). thus f(x^18) = (y^n)^18 = y^(18n) = (y^18)^n = e^n = e, no matter *what* n is to be. this means = {e, x^6, x^12, x^18} will always get mapped by f to the identity element of Z18. so this means f(x) = y cannot be a homomorphism, since y^6 is not the identity of Z18. similarly, f(x) = y^2 cannot be a homomorphism, since y^12 ? e. in fact, for f(x) = y^n to be a homomorphism, 6n has to be a multiple of 18, so n must be a multiple of 3. convince yourself that f(x) = y^3, f(x) = y^6, f(x) = y^9 and f(x) = e (= y^18) all yield homomorphisms, and that there are no other possibilities. 2. if f must be onto (assuming you meant Z12), f(1) must be a generator of Z12, so the possibilities are: f(1) = [1], [5], [7], or [11] 3. in Z3, this is easy: [0]^3 = [0]*[0]*[0] = [0] [1]^3 = [1]*[1]*[1] = [1] [2]^3 = [2]*[2]*[2] = [4]*[2] = [8] = [2], so f3 is just the identity map, which is always a homomorphism. in Z5: [0]^5 = [0] [1]^5 = [1] [2]^5 = [32] = [2] [3]^5 = [243] = [3] [4]^5 = [1024] = [4] so f5 is also just the identity map. in Z4: [0]^4 = [0] [1]^4 = [1] [2]^4 = [16] = [0] [3]^4 = [81] = [1] note that f4([1] + [1]) = f4([2]) = [0] ? [2] = [1] + [1] = f4([1]) + f4([1]), so f4 is not a homomorphism. 4. here, we need to show that if not both of a,b or c,d are 0: f((a+bi)(c+di)) = f(a+bi)f(c+di) now (a+bi)(c+di) = ac - bd + (ad + bc)i, so f((a+bi)(c+di)) = (ac - bd)^2 + (ad + bc)^2 = a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2 = a^2(c^2 + d^2) + b^2(c^2 + d^2) = (a^2 + b^2)(c^2 + d^2) = f(a+bi)f(c+di), so f is a homomorphism. (it should be obvious that (a^2 + b^2)(c^2 + d^2) is non-zero).Related Questions
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