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Unformatted text preview: THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 7 2008 1. (for general discussion) Describe the left and right cosets of the trivial subgroup { 1 } of a group G . In what sense is G/ { 1 } the ‘same’ as G ? In what sense is the additive group of R the ‘same’ as the multiplicative group of R + ? Solution The left and right cosets of { 1 } containing g ∈ G are both equal to { g } . The group G/ { 1 } is isomorphic to G by the map { g } mapsto→ g for g ∈ G . The group ( R , +) is isomorphic to the group ( R + , · ) by the map x mapsto→ e x for x ∈ R . 2. Verify that a group homomorphism is injective if and only if its kernel is trivial. Solution Let φ : G → H be a group homomorphism and put K = ker φ . If φ is injective then, for any k ∈ K , kφ = 1 = 1 φ , since homomorphisms preserve identity elements, so that k = 1, by injectivity, proving K is trivial. Conversely, suppose K is trivial. Let x,y ∈ G such that xφ = yφ . Then, by properties of homomorphisms, ( xy − 1 ) φ = ( xφ )( y − 1 ) φ = ( xφ )( yφ ) − 1 = ( xφ )( xφ ) − 1 = 1 , so that xy − 1 ∈ K = { 1 } , whence xy − 1 = 1, so that x = y . This proves φ is injective. 3. Verify that conjugacy is an equivalence relation on a group. Describe the phenomenon of diagonalisation in linear algebra in terms of conjugacy. Solution Let G be a group. If g ∈ G then g = g 1 , so that g is conjugate to itself, verifying that conjugacy is reflexive. If g,h ∈ G and g is conjugate to h , then g = h k = k − 1 hk for some k ∈ G , so that h = kgk − 1 = ( k − 1 ) − 1 gk − 1 = g k 1 , whence h is conjugate to g , verifying that conjugacy is symmetric. If g,h,k ∈ G such that g is conjugate to h and h is conjugate to k , then there exist x,y ∈ G such that g = h x and h = k y , so that g = x − 1 hx = x − 1 ( y − 1 ky ) x = ( yx ) − 1 k ( yx ) = k yx , whence g is conjugate to k , verifying that conjugacy is transitive. To say that a matrix M is diagonal isable is to say that M is conjugate to a diagonal matrix. 4. Verify that all cyclic groups and all groups of order up to and including five are abelian. Solution If G = ( g ) is cyclic with generator g then, for x,y ∈ G we have x = g i and y = g j for some integers i and j , so that xy = g i g j = g i + j = g j g i = yx , whence G is abelian. By an earlier tutorial exercise, if G has prime order then G is cyclic, so G is abelian by what we have just proved. The trivial group is clearly abelian, so, since 2, 3 and 5 are primes, it remains to consider a group G of order 4 which is not cyclic. If x ∈ G then, by Lagrange’s Theorem, the order of x must be 1, 2 or 4. But the last alternative is excluded sincemust be 1, 2 or 4....
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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, Sets

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