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HW1_Solutions

# HW1_Solutions - MA 541 Modern Algebra I Solutions to HW#1 1...

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MA 541 – Modern Algebra I Solutions to HW #1 1. Which of the following are groups under the given operation? Justify your answers. (a) The collection of odd integers under +. This is not a group, since it is not closed. Consider any element m . Then m + m = 2 m , which is even. (b) The collection of even integers under +. This is a group. Consider any pair of elements m, n 2 Z . Then m = 2 x and n = 2 y for some pair of integers x and y . Therefore, m + n = (2 x ) + (2 y ) = 2( x + y ), which is even, so the operation is binary. Consider elements a , b , and c in 2 Z . Then, by associativity of addition in the integers, ( a + b )+ c = a +( b + c ), so the operation is associative. Consider any element a 2 Z . Then 0 + a = a + 0 = a , so there is an identity, namely 0. Consider any element a 2 Z . Then a has an inverse, namely - a , since a +( - a ) = 0, which we showed above to be the identity element. (c) Z 7 under multiplication. This is not a group, since the element 0 does not have an inverse. For any a ( Z 7 , · ), 0 · a = 0 6 = 1. (d) Z 7 - { 0 } under multiplication (that is, Z 7 with zero removed). This is a group. Consider elements a , b , and c in Z 7 -{ 0 } . Then, by associativity of multiplication in the integers, ( a · b ) · c = a · ( b · c ), so the operation is associative.

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