SOLUTIONS MANUAL
ABSTRACT ALGEBRA, A FIRST COURSE
2ND EDITION
CHAPTER NO. 01: 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:
r3sr4sr = r3s(r4s)r = r3s2r−4r = r31r−3 = 1.
,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)
,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):
(a) (c)
(b)
(d) (f)
(e)
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:
, Exercise: 11 Section 1.1
Question: Sketch a pattern/shape (possibly a commonly known logo) that has rotational symmetry of angle
π
2 but does not have full D4 symmetry.
Solution: The logo for Chase Bank has the desired symmetry.
Exercise: 12 Section 1.1
Question: Consider a regular tetrahedron. We call a rigid motion of the tetrahedron any rotation or composi-
tion of rotations in R3 that map a regular tetrahedron back into itself, though possibly changing specific vertices,
edges, and faces. Rigid motions of solids do not include reflections through a plane.
a) Prove that there are exactly 12 rigid motions of the tetrahedron. Call this set R.
b) Using a labeling of the tetrahedron, explicitly list all rigid motions of the tetrahedron.
c) Explain why function composition ◦ is a binary operation on R.
Solution:
a) Every rigid motion of a tetrahedron will take a vertex and map it into another vertex. There are 4
possibilities for correspondences. Furthermore, for every function that maps a vertex A to a vertex B, there
are three ways to rotate the tetrahedron around the axis OB, where O is the center of the tetrahedron.
Hence, in total there are 4 × 3 = 12 possible rigid motions of a tetrahedron.
b) Let us label the vertices of the tetrahedron as 1, 2, 3, 4. The rigid motions of the tetrahedron are described
by functions f : {1, 2, 3, 4} → {1, 2, 3, 4}. In fact, these functions will be bijections. We can encode such
functions as 3241 to mean that f (1) = 3, f (2) = 2, f (3) = 4, and f (4) = 1. In this manner of encoding
the rigid motions, a complete list is
R ={1234, 1342, 1423, 3241, 4213, 2431, 4132, 2314, 3124, 2143, 3412, 4321}.
c) The definition of a rigid motion allows for a composition of two rigid motions to again be a rigid motion.
Any face will be mapped to any other face the through composition of to rigid motions.
Exercise: 13 Section 1.1
Question: Consider the diagram S ′ below. Sketch the diagram S that has S ′ as a fundamental region with a)
D4 symmetry; b) only rotational square symmetry. [Assume reflection through the x-axis is one of the reflections.]
S′
ABSTRACT ALGEBRA, A FIRST COURSE
2ND EDITION
CHAPTER NO. 01: 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:
r3sr4sr = r3s(r4s)r = r3s2r−4r = r31r−3 = 1.
,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)
,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):
(a) (c)
(b)
(d) (f)
(e)
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:
, Exercise: 11 Section 1.1
Question: Sketch a pattern/shape (possibly a commonly known logo) that has rotational symmetry of angle
π
2 but does not have full D4 symmetry.
Solution: The logo for Chase Bank has the desired symmetry.
Exercise: 12 Section 1.1
Question: Consider a regular tetrahedron. We call a rigid motion of the tetrahedron any rotation or composi-
tion of rotations in R3 that map a regular tetrahedron back into itself, though possibly changing specific vertices,
edges, and faces. Rigid motions of solids do not include reflections through a plane.
a) Prove that there are exactly 12 rigid motions of the tetrahedron. Call this set R.
b) Using a labeling of the tetrahedron, explicitly list all rigid motions of the tetrahedron.
c) Explain why function composition ◦ is a binary operation on R.
Solution:
a) Every rigid motion of a tetrahedron will take a vertex and map it into another vertex. There are 4
possibilities for correspondences. Furthermore, for every function that maps a vertex A to a vertex B, there
are three ways to rotate the tetrahedron around the axis OB, where O is the center of the tetrahedron.
Hence, in total there are 4 × 3 = 12 possible rigid motions of a tetrahedron.
b) Let us label the vertices of the tetrahedron as 1, 2, 3, 4. The rigid motions of the tetrahedron are described
by functions f : {1, 2, 3, 4} → {1, 2, 3, 4}. In fact, these functions will be bijections. We can encode such
functions as 3241 to mean that f (1) = 3, f (2) = 2, f (3) = 4, and f (4) = 1. In this manner of encoding
the rigid motions, a complete list is
R ={1234, 1342, 1423, 3241, 4213, 2431, 4132, 2314, 3124, 2143, 3412, 4321}.
c) The definition of a rigid motion allows for a composition of two rigid motions to again be a rigid motion.
Any face will be mapped to any other face the through composition of to rigid motions.
Exercise: 13 Section 1.1
Question: Consider the diagram S ′ below. Sketch the diagram S that has S ′ as a fundamental region with a)
D4 symmetry; b) only rotational square symmetry. [Assume reflection through the x-axis is one of the reflections.]
S′
