| (3 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| + | =[[HW3_MA453Fall2008walther|HW3]], Chapter 4, Problem 61, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]= | ||
| + | ==Problem Statement== | ||
| + | ''Could somebody please state the problem?'' | ||
| + | |||
| + | ---- | ||
| + | ==Discussion== | ||
The back of the book gives an answer, but I don't find it helpful. Does anyone have a good explaination on how to work this problem? | The back of the book gives an answer, but I don't find it helpful. Does anyone have a good explaination on how to work this problem? | ||
| Line 8: | Line 14: | ||
Corollary 3 says an integer k in <math>Z_n</math> is a generator of <math>Z_n</math> if and only if gcd(n,k) =1 | Corollary 3 says an integer k in <math>Z_n</math> is a generator of <math>Z_n</math> if and only if gcd(n,k) =1 | ||
| + | |||
| + | We have a group G = <a>. Say that a is a generator of the group. The problem says that p be prime. So as a result we know that p and <math>p^n - 1</math> are relativey prime. By the definition of relatively prime we know that the gcd of p and <math>p^n - 1</math> is 1. Therefore, by Corollary 2 we know that G = <math><a^k></math> and that it also generates the group. | ||
| + | Note: <math>(a^p)^k = (a^k)^p</math> | ||
| + | |||
| + | --[[User:Robertsr|Robertsr]] 19:14, 20 September 2008 (UTC) | ||
| + | ---- | ||
| + | [[HW3_MA453Fall2008walther|Back to HW3]] | ||
| + | |||
| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]] | ||
Latest revision as of 17:09, 22 October 2010
HW3, Chapter 4, Problem 61, MA453, Fall 2008, Prof. Walther
Problem Statement
Could somebody please state the problem?
Discussion
The back of the book gives an answer, but I don't find it helpful. Does anyone have a good explaination on how to work this problem?
--Akcooper 16:34, 17 September 2008 (UTC)
Thm. 4.2 says that Let a be an element of order n in a group and let k be a positive integer. Then $ <a^k> = <a^{gcd(n,k)}> $ and $ |a^k| = n/gcd(n,k) $.
Corollary 2 says Let G = <a> be a cyclic group of order n. Then G = $ <a^k> $ if and only if gcd(n,k)=1
Corollary 3 says an integer k in $ Z_n $ is a generator of $ Z_n $ if and only if gcd(n,k) =1
We have a group G = <a>. Say that a is a generator of the group. The problem says that p be prime. So as a result we know that p and $ p^n - 1 $ are relativey prime. By the definition of relatively prime we know that the gcd of p and $ p^n - 1 $ is 1. Therefore, by Corollary 2 we know that G = $ <a^k> $ and that it also generates the group. Note: $ (a^p)^k = (a^k)^p $
--Robertsr 19:14, 20 September 2008 (UTC)
