week07tutsols

# week07tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

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

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

View Full Document
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 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....
View Full Document

## 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.

### Page1 / 4

week07tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

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

View Full Document
Ask a homework question - tutors are online