Show f(x)=x^3 is 1-1 for U(16)

[1800bigk]

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 I am stuck there. Once I show its 1 to 1 I am pretty much done because its guaranteed to be onto since U(16) is finite.

 

[HallsofIvy]

Showing that f(x)= f(y) only if x= y by calculating f(x) for every possible x is tedious but completely valid.

 

[AKG]

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.

 

[1800bigk]

thanks, if a,b in U(16) does (a^3)mod16 = (b^3)mod16 imply (a^3)=(b^3)?