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Chapter 1
Groups
1.1
The Definition of Group
Notes:
1. Show that addition and multiplication modulo m is associative.
2. Show that matrix multiplication is associative.
Problem 1 (Gallian Exercise 1 Page 52). Give two reasons why the set of odd integers under addition is
not a group.
1. The set does not contain an identity under addition.
2. The set is not closed; for example 1 + 1 = 2, but 2 is not an element of the set.
Problem 2 (Gallian Exercise 2 Page 52). Verify the assertion that subtraction of integers is not associative.
For subtraction of integers to be associative we must have that
a − (b − c) = (a − b) − c
for all a, b, c ∈ Z. But, from the definition of subtraction
a − (b − c) = a + (−(b + (−c))) = a + (−b) + c
and
(a − b) − c = a + (−b) + (−c)
Thus a − (b − c) = (a − b) − c ⇔ c = 0. Hence, subtraction of integers is not associative.
Problem 3 (Gallian Exercise 3 Page 52). Show that {1, 2, 3} under multiplication modulo 4 is not a group
but that {1, 2, 3, 4} under multiplication modulo 5 is a group.
Since 2 × 2 = 0 under multiplication modulo 4 it is clear that {1, 2, 3} is not closed under multiplication
modulo 4. For the set {1, 2, 3, 4} under multiplication modulo 5:
1. Closure: The set is closed under multiplication modulo 5.
2. Associativity: Modulo multiplication is associative.
3. Identity: Since 1 × a = a × 1 = a for any a ∈ {1, 2, 3, 4}, the identity is 1.
4. Inverses: 1 and 4 are their own inverses, 2 and 3 are inverses of each other.
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Hence, {1, 2, 3, 4} is a group under multiplication modulo 5.
Problem 4 (Gallian Exercise 4 Page 52). Show that the group GL(2, R) is non-Abelian by exhibiting a pair
of matrices A and B in GL(2, R) such that AB 6= BA.
For example
AB =
1
1
0 0 1
0
=
0 0 1
0
1
,
1
BA =
0 1
0 1
1
1
0
1 0
=
0
1 0
Problem 5 (Gallian Exercise 5 Page 52). Find the inverse of the element
2 6
A=
3 5
in GL(2, Z11 ).
We have the following
A
−1
2
=
3
−1
1
1 5 −6
6
5 −6
5 −6
−1
=
=
=3
5
−3 2
det(A) −3 2
3 −3 2
Now, 3−1 is the multiplicative inverse of 3 in Z11 . Since 3 · 4 = 1, we have 3−1 ≡ 4. Also, −6 ≡ 5 and
−3 ≡ 8. Thus
−1
2 6
5 −6
5 5
20 20
9 9
−1
−1
A =
=4
=3
=
≡
3 5
−3 2
8 2
32 8
10 8
Problem 6 (Gallian Exercise 6 Page 52). TODO...
Problem 7 (Gallian Exercise 7 Page 52). Translate each of the following multiplicative expressions into its
additive counterpart. Assume that the operation is commutative.
1. a2 b3
2. a−2 (b−1 c)2
3. (ab2 )−3 c2 = e
Written additively, × becomes +, an becomes na, and e = 1:
1. 2a + 3b
2. −2a + 2(−b + c)
3. −3(a + 2b) + 2c = 0
Problem 8 (Gallian Exercise 8 Page 52). Show that the set {5, 15, 25, 35} is a group under multiplication
modulo 40. What is the identity element of this group?
Consider the Cayley table:
5
15
25
35
5
25
35
5
15
15
35
25
15
5
Then
1. The set is closed under multiplication modulo 40.
2
25
5
15
25
35
35
15
5
35
25
2. Modulo multiplication is associative.
3. The identity is 25.
4. Each element is its own inverse.
Hence {5, 15, 25, 35} is a group under multiplication modulo 40. Next, consider U (8) = {1, 3, 5, 7} which has
been rearranged on the right:
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
⇒
3
7
1
5
3
1
5
3
7
7
5
1
7
3
1
3
7
1
5
5
7
3
5
1
Thus the Cayley tables for {5, 15, 25, 35} and U (8) are equivalent in the sense that we could interchange 5
and 3, 15 and 7, 25 and 1, and 35 and 5.
Problem 9 (Gallian Exercise 9 Page 52). TODO...
Problem 10 (Gallian Exercise 10 Page 52). TODO...
Problem 11 (Gallian Exercise 11 Page 52). Prove that the set of all 2 × 2 matrices with entries from R and
determinant +1 is a group under matrix multiplication.
Let G denote the group of 2 × 2 matrices with entries in R under multiplication. We need to verify the
four axioms in the group definition:
1. Closure: Let A, B ∈ G, then AB is a 2 × 2 matrix. Since det(A) = det(B) = 1, we have det(AB) =
det(A) · det(B) = 1 · 1 = 1. Thus AB ∈ G.
2. Associativity: Matrix multiplication is associative.
3. Identity: The identity is given by
1
e=
0
0
∈G
1
4. Inverses: For any A ∈ G
A−1 =
a
c
−1
b
d
= det(A)
d
−c
−b
d
=
a
−c
−b
∈G
a
Thus, each element of G has an inverse.
Hence, the set of all 2 × 2 matrices with entries from R under matrix multiplication is a group.
Problem 12 (Gallian Exercise 12 Page 52). TODO...
Problem 13 (Gallian Exercise 13 Page 53). An abstract algebra teacher intended to give a typist a list
of nine integers that form a group under multiplication modulo 91. Instead, one of the nine integers was
inadvertantly left out, so that the list appeared as 1, 9, 16, 22, 53, 74, 79, 81. Which integer was left out?
The group should be closed under multiplication modulo 91. Thus, if we begin to complete a Cayley
table of the given integers under multiplication modulo 91, some combination will yield an integer not in the
list. Indeed 9 · 74 = 29. So 29 is the number left out.
Problem 14 (Gallian Exercise 14 Page 53). Let G be a group with the following property: Whenever a, b,
and c belong to G and ab = ca, then b = c. Prove that G is Abelian.
3
Let x, y ∈ G. Then xyx = xyx, and since associativity holds, we have x(yx) = (xy)x. Letting a = x, b =
yx, c = xy we have ab = ca. But, from the hypothesis, this implies that b = c, and so yx = xy. Hence, G is
Abelian.
Problem 15 (Gallian Exercise 15 Page 53). Let a and b be elements of an Abelian group and let n ∈ Z.
Show that (ab)n = an bn . Is this also true for non-Abelian groups.
Firstly, since groups are associative
(ab)n = (ab)(ab)(ab) . . . (ab) = a(ba)(ba) . . . (ba)b
|
{z
}
n
Then, using the commutative property of Abelian groups, ie ba = ab, we have
(ab)n = a(ba)(ba) . . . (ba)b = aa
. . . a} bb
. . . }b
| {z
| {z
n
n n
n
Hence a b .
TODO: Include proof using Mathematical Induction...
Problem 16 (Gallian Exercise 16 Page 53). TODO...
Problem 17 (Gallian Exercise 17 Page 53). Prove that a group G is Abelian if and only if (ab)−1 = b−1 a−1 .
for all a, b ∈ G.
⇒ By the Shoes-Socks Property (ab)−1 = b−1 a−1 . Since G is Abelian it is commutative so that (ab)−1 =
b a = a−1 b−1 .
⇐ Conversely, suppose that (ab)−1 = a−1 b−1 . But, by the Shoes-Socks Property (ab)−1 = b−1 a−1 . Thus
a−1 b−1 = b−1 a−1 . Taking inverses on both sides ab = ba. Hence, G is Abelian.
−1 −1
Problem 18 (Gallian Exercise 18 Page 53). Prove that in a group, (a−1 )−1 = a for all a.
Firstly, by the inverses property of groups (a−1 )(a−1 )−1 = e.
(a )(a−1 )−1 = (a−1 )a so that by cancellation (a−1 )−1 = a.
Also, (a−1 )a = a−1 a = e.
Thus
−1
Problem 19 (Gallian Exercise 19 Page 53). For any elements a and b from a group and any integer n,
prove that (a−1 ba)n = a−1 bn a
Firstly
(a−1 ba)n = (a−1 ba)(a−1 ba) . . . (a−1 ba)
|
{z
}
n
Since groups are associative
(a−1 ba)n = (a−1 ba)(a−1 ba) . . . (a−1 ba) = a−1 b(aa−1 )b(aa−1 ) . . . b(aa−1 ) ba
|
{z
}
{z
}
|
n
n−1
Applying the rules of inverses (aa−1 = 1) and identity (b1 = b), we have
. . . }b a
. . . }b ba = a−1 bb
a−1 b(aa−1 )b(aa−1 ) . . . b(aa−1 ) ba = a−1 bb
| {z
| {z
{z
}
|
n−1
n−1
n
Hence (a−1 ba)n = a−1 bn a.
Problem 20 (Gallian Exercise 20 Page 53). If a1 , a2 , . . . , an belong to a group, what is the inverse of
a1 a2 · · · an ?
4
By the inverses axiom of the definition of group
(a1 a2 · · · an )(a1 a2 · · · an )−1 = e
Then, multiplying on the left by a1−1 and using the associativity, inverse, and identity axioms:
a1−1 (a1 a2 · · · an )(a1 a2 · · · an )−1 = a−1
1 e
−1
(a−1
= a−1
1 a1 )(a2 · · · an )(a1 a2 · · · an )
1
e(a2 · · · an )(a1 a2 · · · an )−1 = a−1
1
(a2 · · · an )(a1 a2 · · · an )−1 = a−1
1
Similarly, multiplying on the left by a2−1
−1
−1
a−1
= a−1
2 (a2 a3 · · · an )(a1 a2 · · · an )
2 a1
−1
−1
(a−1
= a−1
2 a2 )(a3 · · · an )(a1 a2 · · · an )
2 a1
−1
(a3 · · · an )(a1 a2 · · · an )−1 = a−1
2 a1
Continuing this process
−1
(a1 a2 · · · an )−1 = an−1 · · · a−1
2 a1
This is the generalised Shoes-Socks Property.
Problem 21 (Gallian Exercise 21 Page 53). The integers 5 and 15 are among a collection of 12 integers
that form a group under multiplication modulo 56. List all 12.
The group should be closed under the operation of multiplication modulo 56. Starting with 5 · 15 = 19,
and then 5 · 19 = 39, etc. we continue until we are able to fill a Cayley table. The set is given by
{1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45}
Problem 22 (Gallian Exercise 22 Page 53). TODO...
Problem 23 (Gallian Exercise 23 Page 53). TODO...
Problem 24 (Gallian Exercise 24 Page 53). Construct a Cayley table for U (12).
Since U (12) = {1, 5, 7, 11}, we have the Cayley table
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
5
11
11
7
5
1
Problem 25 (Gallian Exercise 25 Page 53). TODO...
Problem 26 (Gallian Exercise 26 Page 54). Prove that if (ab)2 = a2 b2 in a group G, then ab = ba.
Firstly, (ab)2 = a2 b2 ⇒ abab = aabb. Cancelling on the left we have bab = abb, and then cancelling on
the right, we have the final result ba = ab.
Problem 27 (Gallian Exercise 27 Page 54). Let a, b, and c be elements of a group. Solve the equation
axb = c for x. Solve a−1 xa = c for x.
Solving for x:
axb = c ⇒ a−1 axbb−1 = a−1 cb−1 ⇒ x = a1 cb−1
a−1 xa = c ⇒ aa−1 xaa−1 = aca−1 ⇒ x = aca−1
5
Problem 28 (Gallian Exercise 28 Page 54). Prove that the set of all rational numbers of the form 3m 6n ,
where m and n are integers, is a group under multiplication.
Consider the set G of all rational numbers of the form 3m 6n . We must verify each of the axioms in the
definition of group:
1. Closure: Consider 3s 6t for some s, t ∈ Z. Then (3m 6n )(3s 6t ) = 3m+s 6n+t ∈ G since m + s, n + t ∈ Z.
Thus G is closed under multiplication.
2. Associativity: Multiplication is associative.
3. Identity: The identity is 1 = 30 60 .
4. Inverses: Each element 3m 6n ∈ G has an inverse (3m 6n )−1 = 3−m 6−n since −m, −n ∈ Z.
Problem 29 (Gallian Exercise 29 Page 54). TODO...
Problem 30 (Gallian Exercise 30 Page 54). TODO...
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