hw2sol_Math113_Fall2008 - H a Hence H a is a subgroup Hint...

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Homework 2, selected solutions Math 113: Introduction to Abstract Algebra (Sections 2, 4) 5.51 . (1) H a is closed : let x, y H a . Then xa = ax and ya = ay . Then, by the associative property of groups, we have that ( xy ) a = x ( ya ) = x ( ay ) = ( xa ) y = ( ax ) y = a ( xy ). (2) Because ea = ae = a , we have that e H a . (3) Let x H a . From xa = ax we obtain that xax - 1 = a and then ax - 1 = x - 1 a , showing that x -
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Unformatted text preview: H a . Hence H a is a subgroup. Hint for 6.48 . Every group is a union of its cyclic subgroups, because every element that generates a cyclic group contains that element. None of these cyclic subgroups can be infinite, because otherwise it would be isomorphic to ( Z , +)....
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