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U(16) automorphism

  1. Oct 15, 2005 #1
    I need to show that f(x)=x^3 is an automorphism of U(16) ie.
    {1,3,5,7,9,11,13,15} with operation (multiplication)mod 16. I am having trouble showing that f is 1 to 1. I know it is 1 to 1 because I took each element calculated it to make sure, but how do I show that it is 1 to 1. I would usually assume f(a)=f(b) then show a = b but im stuck there. Once I show its 1 to 1 im pretty much done because its guaranteed to be onto since U(16) is finite.
  2. jcsd
  3. Oct 15, 2005 #2


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    Showing that f(x)= f(y) only if x= y by calculating f(x) for every possible x is tedious but completely valid.
  4. Oct 15, 2005 #3


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    Well if you calculated them all, and they're all different, that constitutes a proof, albeit a rather unelegant one. Suppose x and y are distinct. To show f(x) and f(y) are distinct, observe:

    f(x) - f(y) = x³ - y³ = (x-y)(x² + xy + y²) = (A)(B)

    i.e. A = x-y, B = x² + xy + y²

    Now for f(x) and f(y) to be distinct elements of U(16), AB will have to be divisible by 16. But note that B is an odd number, since it is the sum of 3 numbers, each of which is odd because each of them is a product of two odd numbers. So A would itself would have to be divisible by 16 (possibly being 0). But A is not zero because we're trying to prove that when x and y are distinct, then f maps them to different elements. And A is certainly no other multiple of 16, because two distinct elements of that set can't possibly differ by 16.
  5. Oct 15, 2005 #4
    thanks, if a,b in U(16) does (a^3)mod16 = (b^3)mod16 imply (a^3)=(b^3)?
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