practice2a - MATH 4107 PRACTICE PROBLEMS WITH SOLUTIONS...

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Unformatted text preview: MATH 4107 PRACTICE PROBLEMS WITH SOLUTIONS January 31, 2007 Here are a few practice problems on groups. Try to work these WITHOUT looking at the solutions! After you write your own solution, you can compare to my solution. Your solution does not need to be identical—there are often many ways to solve a problem—but it does need to be CORRECT. 1. Show that if A is a subgroup of a group G and g ∈ G , then gAg- 1 = { gag- 1 : a ∈ A } is a subgroup of G . Solution Let A be a subgroup of G and let g ∈ G be a fixed element. We must show that gAg- 1 is a subgroup of G . Note that gAg- 1 is nonempty since e = geg- 1 ∈ gAg- 1 . To show that gAg − 1 is closed under compositions, suppose that x , y are any two elements of gAg- 1 . By definition, this means that x = gag- 1 and y = gbg- 1 for some a , b ∈ A . Since ab ∈ A , we therefore have that xy = gag- 1 gbg- 1 = g ( ab ) g- 1 ∈ gAg- 1 . To show that gAg- 1 is closed under inverses, suppose that x ∈ gAg- 1 . Then x = gag- 1 for some element a ∈ A . Since a- 1 ∈ A , we therefore have that x- 1 = ( gag- 1 )- 1 = ( g- 1 )- 1 a- 1 g- 1 = ga- 1 g- 1 ∈ gAg- 1 . square 2. Let ϕ : G → H be a homomorphism of groups with kernel N . For a , x ∈ G , show that the following statements are equivalent (i.e., each one implies the other). i. ϕ ( a ) = ϕ ( x ). ii. a- 1 x ∈ N . iii. aN = xN . Solution i ⇒ ii. Suppose that ϕ ( a ) = ϕ ( x ). Then ϕ ( a- 1 x ) = ϕ ( a )- 1 ϕ ( x ) = e , so a- 1 x ∈ ker( ϕ ) = N . ii ⇒ iii. Suppose that a- 1 x ∈ N . Then n = a- 1 x ∈ N . Suppose that z ∈ aN . Then z = am for some m ∈ N . Therefore z = xx- 1 am = xn- 1 m ∈ xN since n- 1 m ∈ N . Thus aN ⊆ xN , and the reverse inclusion is similar. iii ⇒ ii. Suppose that aN = xN . Then a = ae ∈ aN = xN , so a = xn for some n ∈ N . Therefore n = x- 1 a , so a- 1 x = n- 1 ∈ N . iii ⇒ i. Your turn, find the proof. square 3. Let G be a group. Define a ∼ b if there exists a g ∈ G such that b = gag- 1 . Prove that ∼ is an equivalence relation. 1 2 Solution a = aaa- 1 so a ∼ a . If a ∼ b then b = gag- 1 for some g ∈ G . Therefore a = g- 1 bg = g- 1 b ( g- 1 )- 1 . Since g- 1 ∈ G , we have b ∼ a . If a ∼ b and b ∼ c then b = gag- 1 and c = hbh- 1 for some g , h ∈ H . Therefore c = hbh- 1 = hgag- 1 h- 1 = ( hg ) a ( hg )- 1 . Since hg ∈ G , we have a ∼ c . square 4. Let x ∈ R 3 be a fixed vector. The stabilizer of x is the set G x = { A : A is an invertible 3 × 3 matrix and Ax = x } . (a) Prove that G x is a group if the group operation is matrix multiplication. Solution G x is closed under matrix multiplication . Suppose that A , B ∈ G x . Then A and B are invertible 3 × 3 matrices which have the property that Ax = x and Bx = x . The product of invertible matrices is invertible, so AB is an invertible 3 × 3 matrix. Further, ( AB ) x = A ( Bx ) = Ax = x, so AB ∈ G x ....
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This note was uploaded on 08/25/2011 for the course MATH 4107 taught by Professor Staff during the Spring '10 term at University of Florida.

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practice2a - MATH 4107 PRACTICE PROBLEMS WITH SOLUTIONS...

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