Latest revision as of 16:04, 22 October 2010

HW3, Chapter 3, Problem 9, MA453, Fall 2008, Prof. Walther

Problem Statement

Show that U(20) does not equal <k> for any k in U(20). [Hence U(20) is not cyclic.]

Discussion

The problem says show that U(20) does not equal <k> for any k in U(20). [Hence U(20) is not cyclic.] I was trying to understand Example 1 from the Chapter 3 in the text book. where it discusses U(15). I am completely confused about what it is talking about:

U(15) = {1, 2, 4, 7, 8, 11, 13, 14 } Then it goes on to say to find the order of element 7, so |7| = 4 $ 7^1 = 7space 7^2 =4 space 7^3 = 13 space 7^4 = 1 $

Let U(a) = X where X is a group with several elements. Let Z = b + n*a, so all elements in X will satisfy gcd(a,b) = 1.

So for U(15) we'll get {1, 2, 4, 7, 8, 11, 13, 14}

To find an order of an element, y in X, we just have to find a power of the modulo where it will repeat itself. So

$ 7^0 = 1 $, the remainder when 15|1 = 1

$ 7^1 = 7 $, remainder = 7

$ 7^2 = 49 $, remainder of 15|49 = 4

$ 7^3 = 343 $, remainder of 15|343 = 13

$ 7^4 = 2401 $, remainder of 15|2401 = 1

The remainder of $ 7^4 $ is the same as $ 7^0 $, and so we can end here since it the remainders will repeat itself So, the order of |7| = 4 since it has to go through 4 numbers before it repeats itself again.

To make the calculation easier, we can see that $ 7^2 = 49 = (15*n + 4) $ $ 7^3 = (15*n + 4) * 7 $ We know that $ (15*n)*7 $ is going to be divisible by 15, so we just have to find out what the remainder of $ 15|4*7 $.

I just think it's funny that the back of the book tells us to "brute force" to solve this problem. I don't think I've ever used brute force in mathematics before.  ;)

I don't get why U(20) does not equal <k> for any k in U(20). Doesn't U(20) = <1> <3> <7> <9> <11> <13> <17> <19>

check the orders

Isn't this just the example from class?

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