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solutions06 - Problem Set#6 Solutions Chapter 5 1.2 Let G...

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Problem Set #6 Solutions Chapter 5. 1.2. Let G = { 1 , x, x 2 } be a cyclic group of order 3 and let R = Z 2 . Write out addition and multiplication tables for the group algebra RG . We give a partial addition table. Note RG = { 0 , 1 , x, x 2 , 1 + x, 1 + x 2 , x + x 2 , 1 + x + x 2 } . + 1 x x 2 1 0 x + 1 x 2 + 1 x x + 1 0 x 2 + x x 2 x 2 + 1 x 2 + x 0 · 0 1 x x 2 1 + x 1 + x 2 x + x 2 1 + x + x 2 0 0 0 0 0 0 0 0 0 1 0 1 x x 2 1 + x 1 + x 2 x + x 2 1 + x + x 2 x 0 x x 2 1 x + x 2 x + 1 x 2 + 1 1 + x + x 2 x 2 0 x 2 1 x x 2 + 1 x 2 + x 1 + x 1 + x + x 2 1 + x 0 1 + x x + x 2 x 2 + 1 1 + x 2 x 2 + x x + 1 0 1 + x 2 0 1 + x 2 x + 1 x 2 + 1 x + x 2 1 + x x 2 + 1 0 x + x 2 0 x + x 2 x 2 + x 1 + x x + 1 1 + x 2 x 2 + x 0 1 + x + x 2 0 1 + x + x 2 1 + x + x 2 1 + x + x 2 0 0 0 1 + x + x 2 1.3. If M is a semisimple R -module and 0 = x M show that Rx = 0 . Note that M = 0 since x M and x = 0. Thus M = i M i is a direct sum of simple modules M i and so for some j we may write x = x j + y with 0 = x j M j , y i = j M i . Suppose to the contrary that Rx = 0. First, if Rx j = 0 then since A ( M j ) = R , there must exist a z M j such that Rz = 0. But then z = 0, and so since M j is simple there must exist an s R such that sx j = z , for otherwise M j would have a proper submodule generated by x j which would be nonzero since x j = 0. But then z = sx = 0, a contradiction. Thus it must be that Rx j = 0, say rx j = 0. But then since the sum is direct, 0 = rx j = - ry M j i = j M j = 0 , again a contradiction.
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