Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

XXXXXXXXXXExercise 3 1 Let G be the symmetry group of the regular pentagon. Find all elements of G and find generators and relations for G. *2 Let T be a regular tetrahedron, and let G be the symmetry...

1 answer below »

XXXXXXXXXXExercise 3 1 Let G be the symmetry group of the regular pentagon. Find all elements of G and find generators and relations for G. *2 Let T be a regular tetrahedron, and let G be the symmetry group of T (not allowing reflections). Show that G has 12 elements and draw a diagram to show each of these symmetries. Let s be a rotation of 120o through a line through a vertex and the opposite face, and ? and t be two rotations of 180o through the line joining the midpoint of opposite edges. Show that G = {s a ? b t c : 0 = a = 2, 0 = b = 1, 0 = c = 1} and find generators and relations for G. 3 Find the order of each element of Z10 (under addition) and Z * 7 (under multiplication). 4 (i) Find the inverse of 6 in Z * 11. (Not much calculation needed!) Solve 6x = 9 (mod 11). *(ii) Find the inverse of 33 in Z * 47. (Use the method of finding h, k in the h, k-lemma.) Solve 33x = 20 (mod XXXXXXXXXXWrite down and learn the conditions usually used to prove that a subset H of G is a subgroup. 6 Determine, with justification, whether or not the given sets H are subgroups of the given group G: (i) G = R - {0} under ×, H = {x ? R : x = 1}, (ii) G = R - {0} under ×, H = {2 n : n ? Z}, (iii) G = R - {0} under ×, H = {x ? G : x 2 ? Q}, *(iv) G = GL2(R), H = {A =  a b c d  ? G : ad - bc = 1} *(v) G = GL2(R), H = {A =  a b c d  ? G : a, b, c, d = 0} *(vi) G is any group, H = {g ? G : gx = xg for all x ? G}

Document Preview:

XXXXXXXXXXExercise 3 1 Let G be the symmetry group of the regular pentagon. Find all elements of G and ?nd generators and relations for G. *2 Let T be a regular tetrahedron, and let G be the symmetry group of T (not allowing re?ections). Show that G has 12 elements and draw a diagram to show o each of these symmetries. Let s be a rotation of 120 through a line through a o vertex and the opposite face, and ? and t be two rotations of 180 through the line joining the midpoint of opposite edges. Show that a b c G ={s ? t : 0=a= 2,0=b= 1,0=c= 1} and ?nd generators and relations for G. * 3 Find the order of each element of Z (under addition) and Z (under multi- 10 7 plication). * 4 (i) Find the inverse of 6 in Z . (Not much calculation needed!) Solve 6x= 9 11 (mod 11). * *(ii) Find the inverse of 33 in Z . (Use the method of ?nding h,k in the 47 h,k-lemma.) Solve 33x= 20 (mod 47). 5 Write down and learn the conditions usually used to prove that a subset H of G is a subgroup. 6 Determine, with justi?cation, whether or not the given sets H are subgroups of the given group G: (i) G = R-{0} under×, H ={x?R :x= 1}, n (ii) G = R-{0} under×, H ={2 :n? Z}, 2 (iii) G = R-{0} under×, H ={x?G :x ? Q},   a b *(iv) G =GL (R), H ={A = ?G :ad-bc = 1} 2 c d   a b *(v) G =GL (R), H ={A = ?G :a,b,c,d= 0} 2 c d *(vi) G is any group, H ={g?G :gx =xg for all x?G} (Note: GL (R) denotes the group of invertible 2×2 matrices with real entries.) 2 You are advised to attempt all questions. Please hand in the assessed ques- tions (the questions marked with a *) on Monday 18 February at the lecture immediately after reading week. 1

Answered Same Day Dec 22, 2021

Solution

David answered on Dec 22 2021
128 Votes
1. The symmetries of the regular pentagon is isomorphic to the Dihedral group D5. We can number the
vertices cyclically and we obtain{
1, (1 2 3 4 5), (1 3 5 2 4), (1 4 2 5 3), (1 4 2 5 3), (1 5 4 3 2), (2 5)(3 4), (1 5)(2 4), (1 4)(2 3), (1 3)(4 5), (1 2)(3 5)
}
Since this group is isomorphic to D5 and we know that
D5 = 〈x, y | x5 = y2 = (xy)2 = 1〉
2.
Since rotation is mapped to 3 Cycle by φ and since 3-cycles generate An for n ≥ 3 we have that
φ(Sd(T)) ∼= A4. Note that A4 is the alternating group on 4 symbols, the group of even permutations which
has order |A4| = 4!2 = 12.
3. Z10 = {0̄, 1̄, 2̄, 3̄}. And |1̄| = 4 because 1̄ + 1̄ + 1̄ + 1̄︸ ︷︷ ︸
4 times
= 0, similarly we have |2Ì„| = 2, since 2Ì„ + 2Ì„ = 0, and
|3̄| = 4, since 3̄ + 3̄ + 3̄ + 3̄ = 12 ≡ 0 (mod 4). For Z∗7 is the group which has elements from Z7 such that
1
gcd(a, 7) = 1. And so
∣∣Z∗7∣∣ = φ(7) = 6. So to find the order of element 2̄ we have to solve
2x ≡ 1 (mod 7)
which...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here