Math 400
Due Thurs 1/28
Homework 2 - Material from Chapters 1-2
1. Find all six elements of D3 , the symmetries of an equilateral triangle. (Describe with
pictures and/or words, and give each a short name, like we did with D4 .)
Solution: My starting position has the triangle positioned with one vertex pointing
up. There are six elements of D3 . The “do nothing” or “starting position” element
I’ll call R0 , for rotation by 0 degrees. Then we also have R120 and R240 , where I’m
going to use rotation clockwise. (If you choose to rotate counterclockwise, your R120
and R240 wil be the reverse of mine.
Then there are three “flips” (reflections). I’ll call them V (the reflection across a
vertical line, D (reflection across the line from the bottom right vertex to the midpoint
of the left side), and D0 (reflection across the line from the bottom left vertex to the
midpoint of the right side.)
2. Write out a complete Cayley table (“multiplication table”) for D3 .
Solution: In the following table, I use the above notation for motions, and I view the
Row A Column B location as the element BA, which is “do A first and then B”.
R0
R120
R240
V
D
D0
R0 R120 R240 V
D
D0
R0 R120 R240 V
D
D0
R120 R240 R0
D
D0
V
0
R240 R0 R120 D
V
D
0
V
D
D
R0 R240 R120
0
D
V
D R120 R0 R240
D0
D
V R240 R120 R0
3. In Dn , explain geometrically why . . .
(a) a rotation followed by a rotation will always be a rotation
Solution: Rotations keep the “front” of the polygon facing toward you, so you
end up looking at the “front”, so it’s some kind of rotation.
(b) a reflection followed by a reflection will always be a rotation
Solution: Each reflection reverses the shape: First front to back, then back to
front. So you end up looking at the front, so it’s a rotation.
(c) a rotation and a reflection, in either order, will always be a reflection
Solution: The rotation leaves the orientation (back/front position) unchanged,
and the reflection reverses it. So overall you reverse once and remain unchanged
once, giving a net result that the orientation is reversed. So you end up looking
at the back, so it’s a reflection.
4. Describe the symmetries of a nonsquare rectangle (using pictures and/or words). How
many elements are in the symmetry group of a nonsquare rectangle?
Math 400
Due Thurs 1/28
Solution: You can’t rotate by 90 degrees or flip across a diagonal anymore, so you
just have R0 , R180 (the 180 degree rotation about the center), H (a flip across the line
connecting the midpoints of the short edges), and V (a flip across the line connecting
the midpoints of the long edges). This symmetry group has 4 elements.
5. Find pictures of the Chrysler logo and the Mercedes-Benz logo. What is the symmetry
group of each logo?
Solution: The (old-style) Chrysler logo is symmetrically equivalent to a regular pentagon, so its symmetry group is D5 . The Mercedes-Benz logo is symmetrically equivalent to an equilateral triangle, so its symmetry group is D3 .
6. Give two different reasons why the set of odd integers under addition is not a group.
Solution:
• There is no identity element for this set under addition – the additive identity in
the set of integers is 0, and 0 is not odd.
• This set is not closed under addition, since an odd integer plus an odd integer
produces an even integer, not odd.
7. Show that subtraction is not an associative operation on the set of real numbers.
Solution: In general, if a, b, c ∈ R, we have a − (b − c) = a − b + c, while (a − b) − c =
a − b − c. So the only time they’re equal is if c = 0, which means subtraction isn’t
associative.
Or, you can choose an example, like 2 − (1 − 4) = 5 but (2 − 1) − 4 = −3. Since these
aren’t equal, subtraction isn’t associative.
8. Here are some expressions written in multiplicative notation. If the group operation is
addition, what would these be equivalent to in additive notation?
(a) a3 b4 Solution: 3a + 4b, or you can write it out as a + a + a + b + b + b + b.
(b) a−1 b2 a3 Solution: −a + 2b + a3 , or −a + b + b + a + a + a
(c) (ab3 )3 c−2 = e Solution: a + 3b + a + 3b + a + 3b − 2c = 0. Note that you could
write the beginning part as 3(a + 3b), but that’s NOT the same as 3a + 9b unless
you know the group is Abelian (that is, the operation is commutative).
9. Prove that the set of all 2 × 2 matrices with entries in R and determinant equal to ±1
is a group under matrix multiplication. (If you haven’t taken linear algebra, talk to a
friend for help or come see me for some facts about matrices, multiplication, inverses,
and determinants.)
Proof: We need to prove that this set is closed under matrix multiplication, that
matrix multiplication is associative, and that the set contains an identity element and
an inverse for each element under matrix multiplication.
Note that, from linear algebra, we know that if A and B are 2 × 2 matrices then
det(AB) = det(A) det(B). Thus if each of A and B has a determinant of ±1, the
Math 400
Due Thurs 1/28
determinant of their product AB is (±1)(±1) = ±1, so the product is back in this set.
Thus the set is closed under matrix multiplication.
Also from linear algebra, we know matrix multiplication is always associative (since,
for one thing, it is equivalent to composition of linear transformations, and function
composition is always associative).
Now the identity matrix I has det(I) = 1 so I is in this set, and I is the identity
element for matrix multiplication.
Finally, we know that a matrix A is invertible under matrix multiplication if and only
if det(A) 6= 0, and if it is invertible then det(A−1 = 1/ det(A). So all matrices with
determinant ±1 are invertible, and their inverses have determinant 1/(±1) = ±1, so
the inverse of each set element is in the set.
Therefore, this set forms a group under matrix multiplication. 10. Prove that in a group, (a−1 )−1 = a for all a.
Solution: There are several ways to approach this. Here’s one:
Proof: Let a ∈ G for some group G. Then we know that a−1 ∈ G and
aa−1 = e.
(∗∗)
Now since a−1 ∈ G, it has an inverse in G, which we write as (a−1 )−1 . Multiply both
sides of equation (**), on the right, by the element (a−1 )−1 . We obtain
aa−1 (a−1 )−1 = e (a−1 )−1 .
Since the group operation is associative, we rewrite the left side of this equation as
a a−1 (a−1 )−1 = e (a−1 )−1 .
Now the product in square brackets on the left simplifies to e, since it’s a product of
an element and its inverse. So we have
ae = e (a−1 )−1
and thus
a = (a−1 )−1 .
Another approach is less reliant on formal symbol manipulation and more conceptual:
Alternate Proof: By definition (a−1 )−1 is “the element of G that you combine with
a−1 using the group operation to get e.” But we already know that a itself satisfies
that requirement, since aa−1 = a−1 a = e. Therefore, the element (a−1 )−1 must be
exactly the same thing as the element a. Math 400
Due Thurs 1/28
11. Prove that if (ab)2 = a2 b2 for some a, b in a group, then ab = ba.
Proof: (Here I’m using the cancellation property of groups.) Note (ab)2 means ab
times ab, which is written abab. Then suppose (ab)2 = a2 b2 . Rewriting, this means
abab = aabb. We can cancel an a from the left of both sides of this equation, leaving
bab = abb. Now we can cancel a b from the right of both sides of this equation, leaving
ba = ab, as desired. 12. Suppose the table below is a group table (Cayley table). Fill in the blank spaces. (Here
e is the identity element. No justification is needed.)
Answer:
e
a
b
c
d
e
e
a
b
c
d
a
a
b
c
d
e
b
b
c
d
e
a
c
c
d
e
a
b
d
d
e
a
b
c