1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: U(16) automorphism
  1. Automorphism Group (Replies: 2)

Loading...