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Unformatted text preview: Solutions, Algebra hw, Week 3 by Bracket 3.4. Prove that in any group, an element and its inverse have the same order. Let g be an element in some group G . Suppose that g has finite order n , so g n = e . We multiply on the right by ( g 1 ) n , to obtain g n ( g 1 ) n = ( g 1 ) n . Using the definition of the identity element and and the inverse of an element, we can collapse the left side until we are left with e = ( g 1 ) n , so g 1 has order n as well. Alternatively, suppose that g is of infinite order. Then g 1 must be of infinite order as well. To see this, we merely need note that g is the inverse of g 1 , and thus if g 1 has finite order then g must have finite order from the first part of the proof. 3.6. Let x belong to a group. If x 2 6 = e and x 6 = e , prove that x 4 6 = e and x 5 6 = e . What can we say about the order of x ? Clearly x 6 = e , since otherwise x n = e for all n ∈ Z . If x 4 = e , then x 6 = ( x 2 )( x 4 ) = x 2 e = x 2 6 = e , which is a contradiction. Similarly, ifwhich is a contradiction....
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 Spring '08
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 Algebra, Abelian, OneStep Subgroup Test

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