Show φ is a homomorphism from integers to the group of non-zero complex numbers under multiplication?
Let C^∗ be the group of nonzero complex numbers under multiplication. Let φ : Z→C^∗ be the function defined by φ(n) =e^(2πin/12).
a) Show that φ is a homomorphism.
b) What is Ker φ?
c) Show that there is a unique homomorphism from h: Z/12Z→C^∗ such that h(1 mod 12) = e^(2πi/12).
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a) Suppose m, n are in Z. Then φ(m+n) = e^(2πi(m+n)/12) = e^(2πim/12)*e^(2πin/12) = φ(m)*φ(n). This is the property of group homomorphisms.
b) ker φ = {n in Z | φ(n) = 1} since 1 is the identity in C*.
So when does e^(2πin/12) = 1? Well, e^(2πin/12) = cos(2πin/12) + isin(2πin/12). We need, therefore,
2πin/12 = 2πik, where k = 0, +/-1, +/-2, ...
n = 0, +/-6, +/-12, +/-18, +/-24, ...
Therefore, ker φ = 6Z, i.e., multiples of 6.
c) 1 mod 12 (call it [1], the equivalence class of 1) is a generator for Z/12Z. In other words, <[1]> = Z/12Z. As such, h is uniquely determined by where it maps [1]. We can see this explicitly from the homomorphism property as follows:
h([2]) = h([1]+[1]) = h([1])*h([1]) = e^(4πi/12)
h([3]) = h([1]+[1]+[1]) = h([1])*h([1])*h([1]) = e^(6πi/12)
etc.