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Unformatted text preview: THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 6 2008 1. (for general discussion) Let G be a group. (a) Verify that the relation ρ on G defined by xρy iff x = y or x = y 1 is an equivalence relation. (b) What are the possibilities for  x  if the equivalence class containing x is { x } ? (c) Deduce that if G has even order then G contains an element of order 2. (d) Of which famous theorem have you proved a special case? Solution (a) Certainly xρx , since x = x , so that ρ is reflexive. If xρy then x = y or x = y 1 , so that y = x or y = ( y 1 ) 1 = x 1 , so that y ρx , verifying that ρ is symmetric. Suppose that xρyρz , Then either (i) x = y = z , (ii) x = y and y = z 1 , so that x = z 1 , (iii) x = y 1 and y = z , so that x = y 1 = z 1 , or (iv) x = y 1 and y = z 1 , so that x = y 1 = ( z 1 ) 1 = z . In each case x = z or x = z 1 , so that xρz , verifying transitivity. Thus ρ is an equivalence. (b) If the equivalence class of x is { x } then x = x 1 , so that x 2 = 1, yielding  x  = 1 or 2. (c) Suppose G has even order. If G has no element of order 2 then, by the previous part, all of the ρclasses except for { 1 } contain 2 elements, so that G has an odd number of elements, contradicting that its order is even. Hence G has an element of order 2. (d) This is the special case of Cauchy’s Theorem when the prime is 2. 2. Let φ : G → H be an isomorphism between groups G and H . What is the relationship between  g  and  gφ  for any g ∈ G ? Solution We have  g  =  gφ  for all g ∈ G . To see this, put n =  g  and m =  gφ  . Then g n = 1, so that, by the homormorphic property and the fact that a homomorphism maps the identity element to the identity element, ( gφ ) n = ( g n ) φ = 1 φ = 1 . Hence m ≤ n by minimality of order. Similarly n ≤ m , so that m = n . 3. Verify that isomorphism is an equivalence relation on the class of all groups. Solution Observe that the identity mapping is an isomorphism from any group to itself, so that ∼ = is reflexive. Suppose G ∼ = H and let φ : G → H be an isomorphism. In particular φ is a bijection so has an inverse bijection φ 1 . If x,y ∈ H then, since φ is surjective, x = aφ and y = bφ for some a,b ∈ G , so that ( xy ) φ 1 = (( aφ )( bφ )) φ = (( ab ) φ ) φ 1 = ab = ( xφ )( yφ ) ....
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This note was uploaded on 09/12/2009 for the course MATH 2968 taught by Professor Easdown during the One '09 term at University of Sydney.
 One '09
 easdown
 Algebra

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