oct11 - Math 3124 Tuesday, October 11 October 11, Ungraded...

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Math 3124 Tuesday, October 11 October 11, Ungraded Homework The starting position is a b c d e a b e c d e d We are given bd = e and ec = d . Now which element is the identity? The equality bd = e rules out b and d , and the equality ec = d rules out e and c . We are only left with a , so a is the identity. Therefore we can fill in the a column and a row to obtain a b c d e a a b c d e b b e c c d d e e d Now consider bc . By looking at the b row it cannot be b or e , and by looking at the c column it can’t be c or d . Therefore it must be a . Since bc = a and a is the identity, we conclude that cb = a . We now consider cd . By looking at the d column it can’t be d or e , and by looking at the c row it can’t be c or a . Therefore it must be b , and we obtain the following table. a b c d e a a b c d e b b a e c c a b d d e e d We must now have c 2 = e , and the rest is easy; the final table is a b c d e a a b c d e b b d a e c c c a e b d d d e b c a e e c d a b Problem 17.15 on page 92. Let
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.

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oct11 - Math 3124 Tuesday, October 11 October 11, Ungraded...

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