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:
,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 sr3 s sr sr2
r2 r2 r3 1 r sr2 sr3 s sr
r3 r3 1 r r2 sr sr2 sr3 s (1.1)
s s sr sr2 sr3 1 r r2 r3
sr sr sr2 sr3 s r3 1 r r2
sr2 sr2 sr3 s sr r2 r3 1 r
sr3 sr3 s sr sr2 r r2 r3 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
, 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 2β cos α cos 2β − sin α sin 2β cos α sin 2β + sin α cos 2β
=
sin α cos α sin 2β − 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)
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:
,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 sr3 s sr sr2
r2 r2 r3 1 r sr2 sr3 s sr
r3 r3 1 r r2 sr sr2 sr3 s (1.1)
s s sr sr2 sr3 1 r r2 r3
sr sr sr2 sr3 s r3 1 r r2
sr2 sr2 sr3 s sr r2 r3 1 r
sr3 sr3 s sr sr2 r r2 r3 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
, 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 2β cos α cos 2β − sin α sin 2β cos α sin 2β + sin α cos 2β
=
sin α cos α sin 2β − 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)
