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A U(55) Problem

  • Thread starter e(ho0n3
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  • #1
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Homework Statement
Show that [itex]U(55)^3 = \{x^3 \, : \, x \in U(55)\}[/itex] is U(55).

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

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

Answers and Replies

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

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