A Solutions Manual for Abstract Algebra: A First Course, 2nd. ed.
Author
Stephen Lovett Wheaton College
The author would like to thank all the students who assisted with the creation of this solutions manual. In particular, we would like to thank Joel Stapleton, Caleb DeMoss, Daniel Bradley, Jeffrey Burge, and Daniel Windus. This solutions manual is intended for the use by faculty using Abstract Algebra: A First Course (2nd. ed., by Stephen Lovett from CRC Press, ISBN: 978-1032289397) as a textbook in a course. This solutions manual is not intended for unrestricted distribution. Posting the solutions manual to any website is not allowed. Errors or typos crept into the statements of a few problems. The beginning of each chapter in this solution manual lists a few errata and a few additional hints for the exercises. The author invites faculty who use this textbook to submit suggestions for improvements for a future version and suggestions for improvements to any of the solutions provided in this manual to the following email: stephen.lovett@wheaton.edu
1 | Groups Hints and Corrections Exercise 1.7.14 is simply wrong.
1.1 – Symmetries of a Regular Polygon Exercise: 1 Section 1.1 Question: Use diagrams to describe all the dihedral symmetries of the equilateral triangle. Solution: The equilateral triangle has 6 dihedral symmetries.
identity
rotation 120◦
rotation 240◦
reflection through x-axis
reflection
reflection
Exercise: 2 Section 1.1 Question: Write down the composition table for D4 . Solution: Composition table for D4 where the entries give a ◦ b. a\b 1 r r2 r3 s sr sr2 sr3
1 1 r r2 r3 s sr sr2 sr3
r r r2 r3 1 sr sr2 sr3 s
r2 r2 r3 1 r sr2 sr3 s sr
r3 r3 1 r r2 sr3 s sr sr2
s s sr3 sr2 sr 1 r3 r2 r
sr sr s sr3 sr2 r 1 r3 r2
sr2 sr2 sr s sr3 r2 r 1 r3
sr3 sr3 sr2 sr s r3 r2 r 1
Exercise: 3 Section 1.1 Question: Determine what r3 sr4 sr corresponds to in dihedral symmetry of D8 . Solution: In dihedral symmetry of D8 , we have the following algebraic identities on r and s: r8 = 1,
s2 = 1,
rk s = sr−k .
So for our element, progressively change it to put all the s terms to the left: r3 sr4 sr = r3 s(r4 s)r = r3 s2 r−4 r = r3 1r−3 = 1. 3
(1.1)
4
CHAPTER 1. GROUPS
Exercise: 4 Section 1.1 Question: Determine what sr6 sr5 srs corresponds to as a dihedral symmetry of D9 . Solution: Recall from Corollary 3.5 that srk = rn−k s where in our case n = 9. So, sr6 sr5 srs = sr6 sr5 ssr8 = ssr3 r5 (1)r8 = (1)r8 r8 = r9 r7 = 1r7 = r7 .
Exercise: 5 Section 1.1 Question: Let n be an even integer with n ≥ 4. Prove that in Dn , the element rn/2 satisfies rn/2 w = wrn/2 for all w ∈ Dn . Solution: From the paragraph above Proposition 3.1.4 we can write w ∈ Dn as w = sa rb where a is either 0 or 1. Consider rn/2 sa rb . We have two cases. Case 1: a = 0 So we have rn/2 rb = rn/2+b = rb+n/2 = rb rn/2 . Case 2: a = 1 Now, rn/2 srb = srn−n/2 rb = srn/2 rb = srn/2+b = srb+n/2 = srb rn/2 . In both cases, we see that rn/2 w = wrn/2 . Exercise: 6 Section 1.1 Question: Let n be an arbitrary integer n ≥ 3. Show that an expression of the form ra sb rc sd · · · is a rotation if and only if the sum of the powers on s is even. Solution: For any numbers l and m we have rl sm = sm rl−m . So we can move all powers of s around without changing the exponent’s value. Since we can rewrite any element as sj rk , we have ra sb rc sd · · · = sb+d+··· rm for some m. Now, if b+d+· · · is an even number then sb+d+··· rm = s2 s2 · · · s2 rm = (1)(1) · · · (1)rm = 1rm = rm and our element is a rotation. If b+d+· · · is an odd number then sb+d+··· rm = s1 s2 · · · s2 rm = s(1)(1) · · · (1)rm = srm and our elements is not a rotation. Exercise: 7 Section 1.1 Question: Use linear algebra to prove that Rα ◦ Fβ = Fα/2+β ,
Fα ◦ Rβ = Fα−β/2
, and Fα ◦ Fβ = R2(α−β) .
Solution: As linear transformations on R2 → R2 , the matrices of the rotation Rα and of the reflection Fβ with respect to the standard basis are respectively cos α − sin α cos 2β sin 2β and . sin α cos α sin 2β − cos 2β The matrix for Rα ◦ Fβ is cos α − sin α cos 2β sin α cos α sin 2β
sin 2β − cos 2β
cos α cos 2β − sin α sin 2β cos α sin 2β + sin α cos 2β = sin α cos 2β + cos α sin 2β sin α sin 2β − cos α cos 2β cos(α + 2β) sin(α + 2β) = . sin(α + 2β) − cos(α + 2β)
1.1. SYMMETRIES OF A REGULAR POLYGON
5
This matrix corresponds to the reflection Fα/2+β . The matrix for Fα ◦ Rβ is cos 2α sin 2α cos β − sin β cos 2α cos β + sin 2α sin β − cos 2α sin β + sin 2α cos β = sin 2α − cos 2α sin β cos β sin 2α cos β − cos 2α sin β − sin 2α sin β − cos 2α cos β cos(2α − β) sin(2α − β) = sin(2α − β) − cos(2α − β) This matrix corresponds to the reflection Fα−β/2 . The matrix for Fα ◦ Fβ is cos 2α sin 2α cos 2β sin 2β cos 2α cos 2β + sin 2α sin 2β cos 2α sin 2β − sin 2α cos 2β = sin 2α − cos 2α sin 2β − cos 2β sin 2α cos 2β − cos 2α sin 2β sin 2α sin 2β + cos 2α cos 2β cos(2α − 2β) − sin(2α − 2β) = sin(2α − 2β) cos(2α − 2β) This matrix corresponds to the reflection R2(α−β) . Exercise: 8 Section 1.1 Question: Describe the symmetries of an ellipse with unequal half-axes. Solution: The ellipse with unequal half-axes has 4 symmetries. Supposing that the axes of the ellipse are on the x and y axes, then the ellipse has for symmetries: the identity, reflection through the x axis, reflection through the y axis, and rotation by 180◦ , which is the composition of the two reflections. Exercise: 9 Section 1.1 Question: Determine the set of symmetries for each of the following shapes (ignoring shading):
(c)
(a)
(b)
(d)
(e)
(f)
Solution: a) This shape has square rotational symmetry. b) This shape has triangular dihedral symmetry, D3 . c) This flower shape has dodecahedral dihedral symmetry, D12 . d) This shape has octagonal rotational symmetry. e) This shape has pentagonal dihedral symmetry, D5 . f) This shape has 180◦ degree rotational symmetry. Exercise: 10 Section 1.1 Question: Sketch a pattern/shape (possibly a commonly known logo) that has D8 symmetry but does not have Dn symmetry for n > 8. Solution: Here is an example of D8 symmetry: