
#1
Feb911, 01:14 PM

P: 68

1. The problem statement, all variables and given/known data
For any integer n>2, show that there are at least two elements in U(n) that satisfy x^2 = 1. 2. Relevant equations None 3. The attempt at a solution If the definition of the group U(n) is "the set of all positive integers less than n and relatively prime to n" then the group U(5) has elements {1,2,3,4}. Clearly then, 1^2 = 1. But I can't drum up any way to get any other elements in the group to equal 1 when squared. The set U(n) is a group under modulo n, but I am pretty weak on modular arithmetic, and am trying to spend the day boning up on it. So for now, I don't understand how to find another element such that x^2 = 1. 



#2
Feb911, 01:26 PM

P: 68

I'm starting to figure this one out. I chose the case where n = 5. So the set of elements for U(n) = {1,2,3,4}. And under mod 5, 1^2  1 and 4^2 = 1.
I'm trying other mod n's, and it is looking like the integers which will equal 1 when squared are 1, and the largest integer in the set. I don't know how to generalize this for the arbitrary case though. But I'm trying to screw around with a quadratic equation to see where I can go. Since the largest element in U(n) = n1, I am looking for an mod n such that (n1)^2 = 1. But I'm not getting very far with this line of thought. 



#3
Feb911, 02:10 PM

P: 366

Here's one way to look at it. In the integers, the equation x^{2} = 1 has two solutions, namely 1 and 1. Well, this is actually true in U(n) as well, since U(n) is really the multiplicative group [itex] (\mathbb{Z}/n\mathbb{Z})^{\times} [/itex]. So how do you write 1 in [itex] \mathbb{Z}/n\mathbb{Z} [/itex]?




#4
Feb911, 02:15 PM

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HW Helper
Thanks
P: 25,170

Abstract Algebra: Question About the Elements in U(n)
(n1)^2=n^22n+1. Reduce those three terms mod n.




#5
Feb911, 03:23 PM

P: 68





#6
Feb911, 03:26 PM

Sci Advisor
HW Helper
Thanks
P: 25,170

Two numbers are equal mod n if they differ by an even multiple of n. Can you tell if 0 and n^2 are equal mod n?



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