Solution Manual for Abstract Algebra A First Course (Textbooks in Mathematics) 2nd Edition by Stephen Lovet
,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 r3sr4sr 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, rks = 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.
3
,4 CHAPTER 1. GROUPS
Exercise: 4 Section 1.1
Question: Determine what sr6sr5srs corresponds to as a dihedral symmetry of D9.
Solution: Recall from Corollary 3.5 that srk = rn−ks where in our case n = 9. So,
sr6sr5srs = sr6sr5ssr8
= ssr3r5(1)r8
= (1)r8r8
= r9r7
= 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/2w = wrn/2
for all w ∈ Dn.
Solution: From the paragraph above Proposition 3.1.4 we can write w ∈ Dn as w = sarb where a is either 0
or 1. Consider rn/2sarb. We have two cases.
Case 1: a = 0 So we have rn/2rb = rn/2+b = rb+n/2 = rbrn/2.
Case 2: a = 1 Now, rn/2srb = srn−n/2rb = srn/2rb = srn/2+b = srb+n/2 = srbrn/2.
In both cases, we see that rn/2w = wrn/2.
Exercise: 6 Section 1.1
Question: Let n be an arbitrary integer n ≥ 3. Show that an expression of the form
rasbrcsd · · ·
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 rlsm = smrl−m. So we can move all powers of s around without
changing the exponent’s value. Since we can rewrite any element as sjrk, we have rasbrcs·d· · = sb+d+···rm for
some m. Now, if b+d+· · · is an even number then sb+d+···rm = s2s·2 · · s2rm = (1)(1)· · · (1)rm = 1rm = rm and
our element is a rotation. If b+d+ · · · is an odd number then sb+d+···rm = s1·s·2· s2rm = 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)
Solution Manual for Abstract Algebra A First Course (Textbooks in Mathematics) 2nd Edition by Stephen Lovett
, 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):
(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:
,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 r3sr4sr 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, rks = 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.
3
,4 CHAPTER 1. GROUPS
Exercise: 4 Section 1.1
Question: Determine what sr6sr5srs corresponds to as a dihedral symmetry of D9.
Solution: Recall from Corollary 3.5 that srk = rn−ks where in our case n = 9. So,
sr6sr5srs = sr6sr5ssr8
= ssr3r5(1)r8
= (1)r8r8
= r9r7
= 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/2w = wrn/2
for all w ∈ Dn.
Solution: From the paragraph above Proposition 3.1.4 we can write w ∈ Dn as w = sarb where a is either 0
or 1. Consider rn/2sarb. We have two cases.
Case 1: a = 0 So we have rn/2rb = rn/2+b = rb+n/2 = rbrn/2.
Case 2: a = 1 Now, rn/2srb = srn−n/2rb = srn/2rb = srn/2+b = srb+n/2 = srbrn/2.
In both cases, we see that rn/2w = wrn/2.
Exercise: 6 Section 1.1
Question: Let n be an arbitrary integer n ≥ 3. Show that an expression of the form
rasbrcsd · · ·
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 rlsm = smrl−m. So we can move all powers of s around without
changing the exponent’s value. Since we can rewrite any element as sjrk, we have rasbrcs·d· · = sb+d+···rm for
some m. Now, if b+d+· · · is an even number then sb+d+···rm = s2s·2 · · s2rm = (1)(1)· · · (1)rm = 1rm = rm and
our element is a rotation. If b+d+ · · · is an odd number then sb+d+···rm = s1·s·2· s2rm = 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)
Solution Manual for Abstract Algebra A First Course (Textbooks in Mathematics) 2nd Edition by Stephen Lovett
, 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):
(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:
