# A U(55) Problem

1. Jun 11, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Show that $U(55)^3 = \{x^3 \, : \, x \in U(55)\}$ is U(55).

The attempt at a solution
I wrote a Perl script that computes both U(55) and $U(55)^3$ and they both are equal. However, I want to solve this algebraically. My attempt is to define $f(x) = x^3$ (mod 55) and demonstrate that f is an automorphism. If f is an automorphism, surely U(55) = $U(55)^3$. I'm having trouble showing that f is injective.

Suppose f(a) = f(b) and $a \ne b$ where a, b are in U(55). f(a) = f(b) is equiv. to $a^3 = b^3$ (mod 55). I need is reduce this latter equation to a = b (mod 55) but I'm unable to do so.

2. Jun 11, 2008

### HallsofIvy

Staff Emeritus
Saying a3= b3 (mod 55) is the same as saying a3- b3= (a- b)(a2+ ab+ b2)= 0 (mod 55). Does U55 have zero-divisors? What can you say about a2+ ab+ b2= 0 (mod 55)?

3. Jun 11, 2008

### e(ho0n3

I would say U(55) doesn't have zero divisors for if it did, it would not be a group because of closure. If $a^2 + ab + b^2$ = 0 (mod 55), then $a^2 + ab + b^2$ is a multiple of 55 so it is not in U(55).