MATH 3005 Homework Solution
Han-Bom Moon
Homework 2 Model Solution
Chapter 2.
5. In each case, find the inverse of the element under the given operation.
(a) 13 in Z20
13 + 7 mod 20 = 0 ⇒ −13 = 7 in Z20
(b) 13 in U (14).
132 mod 14 = 169 mod 14 = 12 · 14 + 1 mod 14 = 1 ⇒ 13−1 = 13 in U (14)
(c) n − 1 in U (n) (n > 2)
(n − 1)2 mod n = n2 − 2n + 1 mod n = (n − 2)n + 1 mod n = 1
⇒ (n − 1)−1 = n − 1 in U (n)
(d) 3 − 2i in C∗ , the group of nonzero complex numbers under multiplication.
3 + 2i
3 + 2i
3
2
1
=
=
=
+ i
3 − 2i
(3 − 2i)(3 + 2i)
13
13 13
"
#
2 6
11. Find the inverse of the element
in GL(2, Z11 ).
3 5
(3 − 2i)−1 =
"
2 6
3 5
#−1
1
=
2·5−6·3
"
5 −6
−3 2
#
1
=
−8
"
5 −6
−3 2
#
In Z11 , −8 = 3, −6 = 5, and −3 = 8.
"
#
"
#
"
#
1
1 5 5
5
5
5 −6
=
= 3−1
−8 −3 2
3 8 2
8 2
Finally, 3 · 4 mod 11 = 1 so 3−1 = 4 in Z∗11 . Therefore
"
#
"
# "
# "
#
5 5
5 5
20 20
9 9
−1
3
=4
=
=
.
8 2
8 2
32 8
10 8
You may check the axiom of the inverse directly.
13. Translate each of the following multiplicative expressions into its additive counterpart. Assume that the operation is commutative.
(a) a2 b3
2a + 3b
1
MATH 3005 Homework Solution
Han-Bom Moon
(b) a−2 (b−1 c)2
−2a + 2(−b + c)
(c) (ab2 )−3 c2 = e
−3(a + 2b) + 2c = 0
14. For group elements a, b, and c, express (ab)3 and (ab−2 c)−2 without parentheses.
(ab)3 = ababab
(ab−2 c)−2 = ((ab−2 c)−1 )2 = (((ab−2 )c)−1 )2 = (c−1 (ab−2 )−1 )2 = (c−1 (b−2 )−1 a−1 )2
= (c−1 b2 a−1 )2 = c−1 b2 a−1 c−1 b2 a−1
15. Let G be a group and let H = {x−1 | x ∈ G}. Show that G = H as sets.
Here is a standard technique to show the equality between sets. To show the fact
that H = G as a set, we need to show that H ⊂ G and G ⊂ H. Also to show that
H ⊂ G, we have to prove that for any element a ∈ H, a ∈ G as well.
Step 1. H ⊂ G
If a ∈ H, then a = x−1 for some x ∈ G. Because G is a group, x−1 ∈ G. Thus
a ∈ G and H ⊂ G.
Step 2. G ⊂ H
Let a ∈ G. Then a = (a−1 )−1 , because a · a−1 = e = a−1 a so a satisfies the axiom
for the inverse of a−1 . Therefore a is an inverse of some element of G. Thus a ∈ H
and G ⊂ H.
19. Prove that the set of all 2 × 2 matrices with entries from R and determinant +1 is
a group under matrix multiplication.
Let G be the set of all 2 × 2 matrices with entries from R and determinant +1.
Step 1. Closedness of the binary operation
First of all, we need to show that matrix multiplication is a well-defined binary
operation on G, i.e., for any two A, B ∈ G, AB ∈ G as well. Because
det(AB) = det A · det B = 1 · 1 = 1,
AB ∈ G.
Step 2. Associativity
It comes from the associativity of matrix multiplication. (G is a subset of all 2 × 2
matrices. For all matrices the associativity holds so for a subset G it automatically
holds.)
Step 3. Existence of the identity
"
On the set of 2 × 2 matrices, the multiplicative identity is I :=
so I ∈ G.
2
1 0
0 1
#
. det I = 1
MATH 3005 Homework Solution
Han-Bom Moon
Step 4. Existence of inverses
Let A ∈ G. Because its determinant is nonzero, there is a multiplicative inverse
A−1 in the set of 2 × 2 matrices. We need to show that A−1 ∈ G. From AA−1 = I,
1 = det I = det(AA−1 ) = det A · det A−1 = det A−1
so A−1 ∈ G.
Therefore G is a group. (We discussed this group G = SL(2, R) in our class
already.)
22. Let G be a group with the property that for any x, y, z in the group, xy = zx
implies y = z. Prove that G is Abelian. (“Left-right cancellation” implies commutativity.)
We need to show that for any two elements a, b ∈ G, ab = ba. Let’s take x = b,
y = ab, and z = ba. Then xy = bab = zx. From the condition above, y = z so
ab = ba.
25. Prove that a group G is Abelian if and only if (ab)−1 = a−1 b−1 for all a and b in G.
Step 1. G is Abelian ⇒ (ab)−1 = a−1 b−1 for all a, b ∈ G.
Note that (ab)−1 = b−1 a−1 in general. Because G is Abelian, b−1 a−1 = a−1 b−1 .
Step 2. (ab)−1 = a−1 b−1 for all a, b ∈ G ⇒ G is Abelian.
Take a, b ∈ G. Note that a−1 b−1 = (ba)−1 . Therefore (ab)−1 = (ba)−1 . Thus
ab = ((ab)−1 )−1 = ((ba)−1 )−1 = ba and G is Abelian.
26. Prove that in a group, (a−1 )−1 = a for all a.
This exercise should appear earlier...
From the definition of the inverse of a, aa−1 = e. On the other hand, from the
definition of the inverse of a−1 , (a−1 )−1 a−1 = e. So we have aa−1 = (a−1 )−1 a−1 .
By the right cancellation, we have a = (a−1 )−1 .
32. Construct a Cayley table for U (12).
As a set, U (12) = {1, 5, 7, 11}.
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
5
11
11
7
5
1
35. Let a, b, and c be elements of a group. Solve the equation axb = c for x. Solve
a−1 xa = c for x.
axb = c ⇒ xb = exb = a−1 axb = a−1 c
3
MATH 3005 Homework Solution
Han-Bom Moon
xb = a−1 c ⇒ x = xe = xbb−1 = a−1 cb−1
Therefore x = a−1 cb−1 .
a−1 xa = c ⇒ xa = exa = aa−1 xa = ac
xa = ac ⇒ x = xe = xaa−1 = aca−1
So x = aca−1 .
Note that there is a unique solution for given ‘linear’ equation.
37. Let G be a finite group. Show that the number of elements x of G such that x3 = e
is odd. Show that the number of elements x of G such that x2 6= e is even.
A simple technique to show that the number of elements of a set S is even is to
show that we can pair all elements of S. (More formally, we can think a pairing
on a set S as a bijection map p : S → S such that p(x) 6= x and p(p(x)) = x. Then
we can make pairs {x, p(x)} for all elements x ∈ S.) Of course, to show that the
number of elements of S is odd, take one element x out from S and show that we
can pair all elements of S − {x}.
Let S = {x ∈ G | x3 = e} We want to show that the number of elements of S is
odd. Note that e ∈ S. Let S 0 = S − {e}. We will show that the number of elements
in S 0 is even.
Take x ∈ S 0 and consider x−1 . First of all, because (x−1 )3 = (x3 )−1 = e−1 = e,
x−1 ∈ S. Furthermore, if x−1 = e, then x = e. Because x 6= e, x−1 6= e and
x−1 ∈ S 0 .
We claim that collecting {x, x−1 } for all x ∈ S 0 forms pairs for all x ∈ S 0 . First of
all, this is indeed a pair, because if x = x−1 , then x2 = xx−1 = e and x = xe =
xx2 = x3 = e. Secondly, obviously each element in S 0 is in one of such pairs.
Next, suppose that {x, x−1 } ∩ {y, y −1 } 6= ∅. If x = y, then {x, x−1 } = {y, y −1 }.
If x = y −1 , then because x−1 = (y −1 )−1 = y, {x, x−1 } = {y, y −1 } again. In any
cases, such two pairs {x, x−1 } and {y, y −1 } are disjoint or equal. Therefore we
can divide the set S 0 into pairs. Therefore the number of elements of S 0 is even
and that of S is odd.
For next, now consider the set T = {x ∈ G | x2 6= e}. First of all, if (x−1 )2 = e,
then x2 = (x−2 )−1 = ((x−1 )2 )−1 = e−1 = e so x ∈
/ T . So for x ∈ T , x−1 ∈ T . For
x ∈ T , make a pair {x, x−1 } of elements of T .
We claim that this is indeed a ‘pair’ of elements of T . That is, x 6= x−1 . If x = x−1 ,
then x2 = xx−1 = e so x ∈
/ T . Thus x 6= x−1 and {x, x−1 } is a set of two elements
of T .
Obviously, each element of T is in one of such pairs. Moreover, suppose that
{x, x−1 } ∩ {y, y −1 } 6= ∅. If x = y, then {x, x−1 } = {y, y −1 }. If x = y −1 , because x−1 = (y −1 )−1 = y, {x, x−1 } = {y, y −1 } again. Therefore again, such pairs
{x, x−1 } and {y, y −1 } are disjoint or equal. Hence we are able to divide the set T
into pairs and the number of elements of T is even.
4
MATH 3005 Homework Solution
Han-Bom Moon
41. Suppose F1 and F2 are distinct reflections in a dihedral group Dn . Prove that
F1 F2 6= R0 .
(Simple proof) Note that for any reflection F , F 2 = R0 , the identity. In particular,
F1 F1 = R0 . If F1 F2 = R0 = F1 F1 , then by left cancellation, F1 = F2 .
(Geometric proof) Suppose that F1 is a reflection across a line `1 , and F2 is a
reflection across a line `2 . Because F1 6= F2 , `1 6= `2 . Pick a point P on `2 − `1 .
Then since F2 fixes all points on `2 , F1 F2 (P ) = F1 (P ) 6= P . On the other hand,
R0 (P ) = P , because it is the identity map. Therefore F1 F2 6= R0 .
47. Prove that if G is a group with the property that the square of every element is
the identity, then G is Abelian.
Let a, b ∈ G. From the assumption, we know that a2 = e and b2 = e. Moreover,
abab = (ab)2 = e. Then
ab = aeb = a(abab)b = a2 bab2 = ebae = ba.
So G is Abelian.
51. List the six elements of GL(2, Z2 ). Show that this group is non-Abelian by finding
two elements that do not commute.
Note that because Z2 = {0, 1}, there are only two possible distinct entries.
"
# "
# "
# "
# "
# "
#
1 0
1 0
0 1
0 1
1 1
1 1
,
,
,
,
,
0 1
1 1
1 0
1 1
0 1
1 0
It is not Abelian, because for example,
"
# "
# "
#
1 0
0 1
0 1
·
=
,
1 1
1 0
1 1
5
"
0 1
1 0
# "
·
1 0
1 1
#
"
=
1 1
1 0
#
.