(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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| + | ---- | ||
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| + | More so, remember that with U(30), the group operation 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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| + | ---- | ||
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| + | Thanks for the explanation. | ||
Latest revision as of 07:48, 19 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 operation 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
Thanks for the explanation.
