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Then it goes on to say to find the order of element 7, so |7| = 4 | Then it goes on to say to find the order of element 7, so |7| = 4 | ||
<math>7^1 = 7space 7^2 =4 space 7^3 = 13 space 7^4 = 1</math> | <math>7^1 = 7space 7^2 =4 space 7^3 = 13 space 7^4 = 1</math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <math> 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 </math> | ||
| + | So, the order of |7| = 4 since it has to go through 4 numbers before it repeats itself again. | ||
Revision as of 19:53, 16 September 2008
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.
