Chapter 3

Integers and Abelian groups

The set integers Z = {. . . − 2, −1, 0, 1, . . .} is obtained by adding negative num-

bers to the set of natural numbers. This makes arithmetic easier.
Addition satisfies the rules (1.1), (1.2), (1.3) as before. In addition, there is
new operation n 7→ −n satisfying

For each n ∈ Z, n + (−n) = 0 (3.1)

The cancellation law becomes redundant as we will see.

We will now abstract this:
Definition 3.1. An abelian group consists of a set A with an associative com-
mutative binary operation ∗ and an identity element e ∈ A satisfying a ∗ e = a
and such that any element a has an inverse a0 which satisfies a ∗ a0 = e.
Abelian groups are everywhere. Here list a few some examples.
Let Q = {a/b | a, b ∈ Z, b 6= 0} be the set of rational numbers, the R the set
of real numbers and C the set of complex numbers.
Example 3.2. The sets Z, Q, R or C with ∗ = + and e = 0 are abelian groups.
Example 3.3. The set Q∗ , (or R∗ or C∗ ) of nonzero rational (or real or com-
plex) numbers with ∗ = · (multiplication) and e = 1 is an abelian group. The
inverse in this case is just the reciprocal.
Example 3.4. Let n be a positive integer. Let Zn = {(a1 , a2 , . . . an )|a1 , . . . an ∈
Z}. We define (a1 , . . . an ) + (b1 , . . . bn ) = (a1 + b1 , . . . an + bn ) and 0 = (0, . . . 0).
Then Zn becomes an abelian group. Z can be replaced by Q, R or C and these
examples are probably familiar from linear algebra.
Example 3.5. Let n be a positive integer, Zn = {0, 1, . . . n − 1}. Arrange these
on the face of a “clock”. We define a new kind of operation ⊕ called addition
mod n. To compute a ⊕ b, we set the “time” to a and then count off b hours.
We’ll give a more precise description later. Unlike the previous examples, this
is a finite abelian group.

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 Often, especially in later sections, we will simply use + instead ⊕ because
it easier to write. We do this in the diagram below:

0=2+2

3=2+1 1=2+3

2=2+0

Here’s the addition table for Z8 .

⊕ 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7
1 1 2 3 4 5 6 7 0
2 2 3 4 5 6 7 0 1
3 3 4 5 6 7 0 1 2
4 4 5 6 7 0 1 2 3
5 5 6 7 0 1 2 3 4
6 6 7 0 1 2 3 4 5
7 7 0 1 2 3 4 5 6
Notice that the table is symmetric (i.e. interchanging rows with columns
gives the same thing). This is because the commutative law holds. The fact
that that 0 is the identity corresponds to the fact that the row corresponding
0 is identical to the top row. There is one more notable feature of this table:
every row contains each of the elements 0, . . . 7 exactly once. A table of elements
with this property is called a latin square. As we will see this is always true for
any abelian group.
We can now define the precise addition law for Zn . Given a, b ∈ Zn , a ⊕ b =
r(a + b, n), where r is the remainder introduced before.
When doing calculations in Maple, we can use the mod operator. For exam-
ple to add 32 ⊕ 12 in Z41 , we just type

32 + 12 mod 41;

Let n be a positive integer, a complex number z is called an nth root of unity

if z n = 1. Let µn be the set of all nth roots of unity. For example, µ2 = {1, −1}
and µ4 = {1, −1, i, −i}.

Example 3.6. µn becomes an abelian group under multiplication

To see that this statement make sense, note that given two elements z 1 , z2 ∈
µn , their product lies in µn since (z1 z2 )n = z1n z2n = 1 and 1/z1 ∈ µn since

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(1/z1 )n = 1. We can describe all the elements of µn with the help of Euler’s
formula:
eiθ = cos θ + i sin θ.

iθ
e

2(n−1)π
Lemma 3.7. µn = {eiθ | θ = 0, 2π 4π
n , n ,... n }
Proof. The equation z n = 1 can have at most n solutions since it has degree n
(we will prove this later on). So it’s enough to verify that all of the elements
on the right are really solutions. Each element is of the form z = eiθ with
θ = 2πk/n with k an integer. Then

z n = einθ = cos(2πk) + i sin(2πk) = 1.

The lemmas says that the elements are equally spaced around the unit circle
of C.
Since eiθ1 eiθ2 = ei(θ1 +θ2 ) , multiplication amounts to adding the angles. This
sounds suspiciously like the previous example. We will see they are essentially
the same.
Lemma 3.8. (Cancellation) Suppose that (A, ∗, e) is an abelian group. Then
a ∗ b = a ∗ c implies b = c.
Proof. By assumption, there exists a0 with a0 ∗ a = a ∗ a0 = e. Therefore

a0 ∗ (a ∗ b) = a0 ∗ (a ∗ c)

(a0 ∗ a) ∗ b = (a0 ∗ a) ∗ c

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 e∗b=e∗c
b = c.

Corollary 3.9. Given a, there is a unique element a0 , called the inverse, such
that a ∗ a0 = e.
Lemma 3.10. The multiplication table
* a1 . . .
a1 . . .
..
.
of any abelian group A = {a1 , a2 , . . .} forms a symmetric latin square.
Proof. The symmetry follows from the commutative law. Suppose that A =
{a1 , a2 , . . .}. Then the ith row of the table consists of ai ∗ a1 , ai ∗ a2 . . .. Given
a ∈ A, the equation a = ai ∗ (a0i ∗ a) shows that a occurs somewhere in this row.
Suppose that it occurs twice, that is ai ∗ aj = ai ∗ ak = a for aj 6= ak . Then this
would contradict the cancellation lemma.
Let (A, ∗, e) be a group. Given a ∈ A and n ∈ Z, define an by

 a ∗ a . . . a ( n times) if n > 0
an = e if n = 0
 0
a ∗ a0 . . . a0 ( −n times) if n < 0
Often the operation on A is written as +, in which case the inverse of a is
usually written as −a, and we write na instead of an . When A = Z, this noth-
ing but the definition of multiplication. It’s possible to prove the associative,
commutative and distributive laws for Z, but we’ll skip this.

3.11 Exercises
1. Let A = {e, a, b} with e, a, b distinct and the following multiplication table:

* e a b
e e a b
a a e b
b b b a

Is A an abelian group? Prove it, or explain what goes wrong.

2. Let A = {e, a} with a 6= e and the following multiplication table:

* e a
e e a
a a e

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 Is A an abelian group? Prove it, or explain what goes wrong.

3. Let (A, ∗, e) be an abelian group. Let a0 denote the inverse of a. Prove

that e0 = e, (a0 )0 = a and (a ∗ b)0 = a0 ∗ b0 .

4. With notation as above, prove that (an )0 = (a0 )n for any natural number
n by induction. This proves (an )−1 = (a−1 )n = a−n as one would hope.

5. Let (A, ∗, e) and (B, ∗, ²) be two abelian groups. Let A × B = {(a, b) | a ∈

A, b ∈ B}. Define (a1 , b1 ) ∗ (a2 , b2 ) = (a1 ∗ a2 , b1 ∗ b2 ) and E = (e, ²). Prove
that (A × B, ∗, E) is an abelian group. This is called the direct product
of A and B. For example Z2 = Z × Z.

6. Write down the multiplication tables for µ2 , µ3 , µ4 and µ5 .

7. An element ω ∈ µn is called a primitive root if any element can be written

as a power of ω. Check that e2πi/5 ∈ µ5 is primitive. Determine all the
others in this group.

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