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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. | 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> | Note: <math>(a^p)^k = (a^k)^p</math> | ||
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| + | --[[User:Robertsr|Robertsr]] 19:14, 20 September 2008 (UTC) | ||
Revision as of 14:14, 20 September 2008
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)
