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=[[HW3_MA453Fall2008walther|HW3]], Chapter 4, Problem 61, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]=
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==Problem Statement==
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''Could somebody please state the problem?''
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==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?
  
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--[[User:Robertsr|Robertsr]] 19:14, 20 September 2008 (UTC)
 
--[[User:Robertsr|Robertsr]] 19:14, 20 September 2008 (UTC)
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[[HW3_MA453Fall2008walther|Back to HW3]]
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[[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]]

Latest revision as of 16:09, 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)


Back to HW3

Back to MA453 Fall 2008 Prof. Walther

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