(New page: Category:MA453Spring2009Walther I used Theorem 10.1.6 from the textbook (6th edition) with g=g'=7 to solve this problem.)
 
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I used Theorem 10.1.6 from the textbook (6th edition) with g=g'=7 to solve this problem.
 
I used Theorem 10.1.6 from the textbook (6th edition) with g=g'=7 to solve this problem.
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More so, remember that with U(30), the group multiplication IS multiplication, unlike with number 24 where it was addition. Thus, with ker = {1,11}, g = 7, g*ker = {7,77} but 77 = 17 in U(30).
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-K. Brumbaugh
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Revision as of 17:05, 18 February 2009


I used Theorem 10.1.6 from the textbook (6th edition) with g=g'=7 to solve this problem.



More so, remember that with U(30), the group multiplication IS multiplication, unlike with number 24 where it was addition. Thus, with ker = {1,11}, g = 7, g*ker = {7,77} but 77 = 17 in U(30). -K. Brumbaugh


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