week06tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 on G defined by xy 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 xx , since x = x , so that is reflexive. If xy then x = y or x = y- 1 , so that y = x or y = ( y- 1 )- 1 = x- 1 , so that y x , verifying that is symmetric. Suppose that xyz , 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 xz , verifying transitivity. Thus 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 -classes 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 Cauchys Theorem when the prime is 2. 2. Let : G H be an isomorphism between groups G and H . What is the relationship between | g | and | g | for any g G ? Solution We have | g | = | g | for all g G . To see this, put n = | g | and m = | g | . Then g n = 1, so that, by the homormorphic property and the fact that a homomorphism maps the identity element to the identity element, ( g ) n = ( g n ) = 1 = 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 : G H be an isomorphism. In particular is a bijection so has an inverse bijection - 1 . If x,y H then, since is surjective, x = a and y = b for some a,b G , so that ( xy ) - 1 = (( a )( b )) = (( ab ) ) - 1 = ab = ( x )( y ) ....
View Full Document

Page1 / 4

week06tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online