Download Binary Operations

* 8 Groups,
with Appendix containing Rings and Fields.
Binary Operations
Definition We say that ∗ is a binary operation on a set S if, and only if,
∀a, b, a ∗ b ∈ S.
Implicit in this definition is the idea that ∗ is closed, or as we often say,
S is closed under ∗. The idea here is that we can give a value to a ∗ b from
the set in which we started.
Example In Chapter 4 we saw that ×8 is closed on {[0]8 , [2]8 , [4]8 , [6]8 } but
not on {[2]8 , [4]8 , [6]8 }. So the set S is important in the definition above.
Examples Seen in the course:
+ on Z, Zm ,
× on Z, Zm or Z∗m ,
◦, composition of permutations. on Sn , the set of permutations on {1, 2, ..., n}.
Example Not seen in the course:
(a, b) ∗ (c, d) = (ac, bc + d) on Z × Z.
Multiplication of n × n matrices with real entries.
Example Subtraction, −, is a binary operation on Z, since ∀m, n ∈ Z, m −
n ∈ Z. But − is not a binary operation on N since 1, 2 ∈ N, yet 1 − 2 ∈
/ N.
A binary operation might satisfy certain properties that we have seen
before (PJE p.18 for real numbers and p.71 for sets).
Definition (i) A binary operation is commutative if,
∀a, b ∈ S, a ∗ b = b ∗ a,
(ii) A binary operation is associative if,
∀a, b, c ∈ S, (a ∗ b) ∗ c = a ∗ (b ∗ c) .
Example (S, ◦), where S is the set of all function Ω → Ω, under composition,
is not commutative, but is associative.
Example Subtraction, −, on Z is neither commutative, (counterexample:
1 − 2 6= 2 − 1) nor associative (counterexample: 1 − (2 − 3) 6= (1 − 2) − 3).
1
Identities and Inverses
Definition Given a set S and binary operation ∗ we say that e ∈ S is an
identity if, for all a ∈ S,
e ∗ a = a and a ∗ e = a.
We have to check both e ∗ a and a ∗ e since we are not assuming that ∗ is
commutative.
Example (i) 0 is the additive identity on R (or Z, N, C, Zm ) since 0 + x =
x + 0 = x for all x ∈ R (or Z, N, C, Zm respectively).
(ii) 1 is the multiplicative identity on R (or Z, N, C, Zm ) since 1 × x =
x × 1 = x for all x ∈ R (or Z, N, C, Zm respectively).
(iii) {[4]20 , [8]20 , [12]20 , [16]20 , ×}
×
[4]20
[8]20
[12]20
[16]20
[4]20
[16]20
[12]20
[8]20
[4]20
[8]20
[12]20
[4]20
[16]20
[8]20
[12]20
[8]20
[16]20
[4]20
[12]20
[16]20
[4]20
[8]20
[12]20
[16]20
Here the identity is [16]20 .
This last example is important, it shows that we get identities different
to 1 and 0!
Note the use of the word “an” in the definition. But
Lemma Suppose that ∗ is a binary operation on a set S and that (S, ∗) has
an identity. The identity is unique.
Proof Suppose that e and f are two identities on S. Then
e = e ∗ f since f is an identity (used here on the right),
= f since e is an identity (used here on the left).
So we could replace the word “an” by “the” in the definition.
If, in the multiplication table for (S, ∗) , we can find an element whose
row (and whose column) is identical to the heading row (respectively heading
column), then we have found the identity.
2
Definition Let S be a set with a binary operation ∗ and an identity element
e ∈ S. We say that an element a ∈ S is invertible if there exists b ∈ S such
that
a ∗ b = e and b ∗ a = e.
We say that b is the inverse of a, and normally write b as a−1 .
Example In (Z6 , ×) the element [2]6 has no inverse. If it had an inverse,
i.e. [b]6 :
[2]6 [b]6 = [1]6 ,
multiply both sides by [3]6 to get
[6]6 [b]6 = [3]6 , i.e. [0]6 = [3]6 ,
a contradiction.
The problem here is that 6 = 2 × 3 is composite. We have got round this
in two ways in this course. First we can look at (Zp , ×) with p prime, when
every non-zero element has an inverse. The second way it to look at (Z∗m , ×)
where we simply throw away all the elements that don’t have an inverse!
If, for an i ∈ S we can look in its row in the multiplication table and find
the identity in column j, say, and find in row j the identity in column i then
i and j are inverse to each other. If we can do this for every i ∈ S then every
element will have an inverse.
Example {[4]20 , [8]20 , [12]20 , [16]20 , ×}
×
[4]20
[8]20
[12]20
[16]20
[4]20
[16]20
[12]20
[8]20
[4]20
[8]20
[12]20
[4]20
[16]20
[8]20
[12]20
[8]20
[16]20
[4]20
[12]20
[16]20
[4]20
[8]20
[12]20
[16]20
Since the identity is [16]20 we note
[4]20 × [4]20 = [16]20 so [4]−1
20 = [4]20 ,
−1
[8]20 × [12]20 = [16]20 so [8]20 = [12]20 ,
[12]20 × [8]20 = [16]20 so [12]−1
20 = [8]20 .
(The inverse of the identity is always itself!)
3
Semigroups
Definition Let (S, ∗) be a non-empty set with a binary operation. Then
(S, ∗) is a semigroup if ∗ is associative.
Definition A semigroup (S, ∗) is called a commutative semigroup if ∗ is
commutative.
Examples (Z, ×) , (N, +) , (Zm , ×) , (Z∗m , ×) are all commutative semigroups
while S, be the set of all maps from X to X with the composition of functions
is not.
Definition Suppose (S, ∗) is a semigroup and let a ∈ S. Define iteratively
the positive powers of a by
a1 = a and an+1 = a ∗ an for all n ≥ 1.
Example This is exactly how we defined positive powers of a permutation.
Note For the theory of semigroups (and later, groups) we think of ∗ as a
form of multiplication. If the operation were additive, written say as ⊕, then
the corresponding definition of powers would be (n + 1) a = a ⊕ na.
Lemma Index Laws
Let (S, ∗) be a semigroup and a ∈ S. Then for all n, m ∈ N,
(i) am ∗ an = am+n ,
(ii) (an )m = amn .
Proof (i) Let n ≥ 1 be given. Use proof by induction for m ≥ 1 to show
that
∀m ≥ 1, am ∗ an = am+n ,
(1)
First consider m = 1. Then
a1 ∗ an = a ∗ an by definition of a1
= an+1 by definition of an+1 .
Thus result holds for m = 1.
Assume result holds for m = k, i.e. ak ∗ an = ak+n . Consider
ak+1 ∗ an = a ∗ ak ∗ an by definition of ak+1 ,
= a ∗ ak ∗ an since ∗ is associative,
= a ∗ ak+n by inductive hypothesis,
= a(k+n)+1 by definition of power,
= a(k+1)+n .
4
Hence result holds for m = k + 1. Thus by induction, (1) holds. Since n was
arbitrary we have shown
∀n ≥ 1, ∀m ≥ 1, am ∗ an = am+n .
as required.
(ii) Let n ≥ 1 be given. Use proof by induction for m ≥ 1 to show that
∀m ≥ 1, (an )m = amn ,
(2)
First consider m = 1. Then
(an )1 = an by definition of first power
= a1×a .
Thus result holds for m = 1.
Assume result holds for m = k, i.e. (an )k = akn . Consider
(an )k+1 =
=
=
=
an ∗ (an )k by definition of k + 1th power,
an ∗ akn , by inductive hypothesis,
an+kn by part (i),
a(k+1)n .
Hence result holds for m = k + 1. Thus by induction, (2) holds. Since n was
arbitrary we have shown
∀n ≥ 1, ∀m ≥ 1, (an )m = amn
as required.
Note this means that in particular,
an+1 = an ∗ a.
This could have been taken as the definition of an+1 .
Remark If the semigroup (S, ∗) is commutative we also have
∀a, b ∈ S, (a ∗ b)n = an ∗ bn .
This need not happen in general. Can you think of an example?
One reason for introducing semigroups is the following.
5
(3)
Lemma Suppose that (S, ∗) is a semigroup and a ∈ S has an inverse. Then
the inverse is unique.
Proof If an element a has two inverses, b and c say, then
b =
=
=
=
b ∗ e = b ∗ (a ∗ c) since c is an inverse of a,
(b ∗ a) ∗ c by associativity,
e ∗ c since b is an inverse of a,
c.
Groups
Definition Given a set G and binary operation ∗ we say that (G, ∗) is a
group if, and only if,
G1 G is closed under ∗,
G2 ∗ is associative on G,
G3 (G, ∗) has an identity element, i.e.
∃e ∈ G : ∀a ∈ G, e ∗ a = a ∗ e = a,
G4 every element of (G, ∗) has an inverse, i.e.
∀a ∈ G, ∃a0 ∈ G : a ∗ a0 = a0 ∗ a = e.
Note that a group is a “semigroup satisfying G3 and G4”. Thus earlier
results hold, namely that the identity element will be unique as will each
inverse. Also for m, n ∈ N and g ∈ G we have g m g n = g m+n and (g n )m = g nm .
Definition For a group (G, ∗) we say it is an infinite group if G is an
infinite set and we say that (G, ∗) is a finite group if G is a finite set. In
the latter case |G| denotes the number of elements in G and is called the
order of (G, ∗).
We say that (G, ∗) is a commutative or abelian group (after Niels
Abel) if, and only if, it is a group and ∗ is commutative.
Example (Z, +) and (Zm , +m ) are additive groups, the first infinite and the
second finite.
Example ({i, −1, −i, 1} , ×), where i2 = −1, is a finite multiplicative group.
Example The set of permutations of a set of n elements, (Sn , ◦), is a (nonabelian) group. In fact, if you look back you can see we called it a symmetric
6
group before we knew what “group” meant. This will be one of our main
examples when illustrating the properties of groups in general. It is a group
because: G1 the composition of bijections is a bijection, G2 composition of
functions is always associative, G3 the permutation that fixes every element
is the identity and G4 every bijection has an inverse.
Question But why do we call (Sn , ◦) the symmetric group? Consider, as an
example, n = 4. Think of a square in the plane, center at the origin, with
vertices labelled clockwise, 1,2,3 and 4. What symmetries does the square
have? It has rotational symmetries about the origin. If we rotate by π/2 in
the clockwise direction we see that corners map 1 → 2, 2 → 3, 3 → 4 and
4 → 1. So this rotation can be represented by the cycle (1, 2, 3, 4).
In the other direction what would (1, 2) ◦ (3, 4) represent? It would be a
reflection in a line through the origin.
For Student: What are the permutations that represent the other symmetries of the square?
In this way we see that S4 contains the symmetries of the square. Hence
the use of the word “symmetry” in the name of (Sn , ◦).
Definition Let (G, ∗) be a group and g ∈ G. Define the non-positive
powers of g by
n
g 0 = e and g −n = g −1 for n ≥ 1.
Example This is exactly how we defined negative powers of a permutation.
(So the earlier, iterative definition should be used for the positive power
h where h = g −1 .)
n
7
Elementary consequences of the axioms:
Theorem Let (G, ∗) be a group.
(i) For a, x, y ∈ G if x ∗ a = y ∗ a then x = y (a cancellation result),
(ii) For a, x, y ∈ G if a ∗ x = a ∗ y then x = y (a cancellation result),
(iii) e−1 = e, where e is the identity,
−1
(iv) For all x ∈ G, (x−1 ) = x,
(v) For all x, y ∈ G, (x ∗ y)−1 = y −1 ∗ x−1 ,
(vi) For x1 , x2 , ..., xn ∈ G,
−1
−1
(x1 ∗ x2 ∗ ... ∗ xn )−1 = x−1
n ∗ ... ∗ x2 ∗ x1 ,
(vii) For m, n ∈ Z and g ∈ G, g m ∗ g n = g m+n and (g n )m = g nm .
Proof not given in course.
Note We have already seen examples of (i) and (ii) for (Z∗m , ×) and permutation groups (Sn , ◦). Result (vii) was known previously in semigroups only
for m, n ∈ N.
End of course in 2008
Definition Let (G, ∗) be a group with identity e and let g ∈ G. If there exists
n such that g n = e then the order of g is the least positive d : g d = e. If
no such n exists we say g has infinite order.
Example This generalises the definition of order for permutations given in
Chapter 6.
Example (i) (Z∗5 , ×) : [2]25 = [4]5 , [2]35 = [3]5 , [2]45 = [1]5 . So the order of [2]5
is 4.
For students, show that the orders of [3]5 and [4]5 are also 4.
(ii) In ({i, −1, −i, 1} , ×) the order of −1 is 2, of i and −i are 4.
(iii) (Z6 , +) is an additive group, identity [0]6 . To find orders we add until
we get zero. So
[2]6 + [2]6 = [4]6 , [2]6 + [4]6 = [6]6 = [0]6 ,
[3]6 + [3]6 = [6]6 = [0]6 .
Thus the order of [2]6 is 3, and the order of [3]6 is 2.
Something for future years: for finite groups the order of an element divides
the order of the group!
8
Multiplication Tables for Finite Groups
Suppose ∗ is a closed binary operation on a finite set S = {a1 , ..., an }.
Then the multiplication table for (S, ∗) is
∗
a1
a2
a1
a1 ∗ a1
a1 ∗ a2
a2
..
.
..
.
..
.
a2 ∗ a1
ai
..
.
···
an
···
···
aj
..
.
..
.
..
.
..
.
..
.
···
···
···
an
···
ai ∗ aj
..
.
..
.
···
···
···
···
Examples of these tables have been seen throughout this course, for example for (S3 , ◦) , (Z5 , +) , (Z8 , ×) , (Z∗5 , ×) and (Z∗8 , ×).
If we can fill in every entry of the table with an element of S then the
operation ∗ is closed.
In a group (G, ∗) we have an identity so, as noted before, in the multiplication table we will find an element whose row (and whose column) is
identical to the heading row (respectively heading column).
Also, in a group every element has an inverse, hence for any i ∈ S we can
look in its row in the multiplication table and find the identity in column j,
say, and find in row j the identity in column i.
Another property of the table of a group is
Lemma If (G, ∗) is a group then every row (and every column) contains each
element of G once.
Proof If in row a two different columns, b and c say, contained the same
element then a ∗ b = a ∗ c. The cancellation rule gives b = c which contradicts
them being different.
Similarly, If in column α two different rows, β and γ say, contained the
same element then β ∗ α = γ ∗ α. The cancellation rule gives β = γ which
contradicts them being different.
9
Example If we look back at
group.
× [0]8 [1]8
[0]8 [0]8 [0]8
[1]8 [0]8 [1]8
[2]8 [0]8 [2]8
[3]8 [0]8 [3]8
[4]8 [0]8 [4]8
[5]8 [0]8 [5]8
[6]8 [0]8 [6]8
[7]8 [0]8 [7]8
the table for (Z8 , ×) we see that it is not a
[2]8
[0]8
[2]8
[4]8
[6]8
[0]8
[2]8
[4]8
[6]8
[3]8
[0]8
[3]8
[6]8
[1]8
[4]8
[7]8
[2]8
[5]8
[4]8
[0]8
[4]8
[0]8
[4]8
[0]8
[4]8
[0]8
[4]8
[5]8
[0]8
[5]8
[2]8
[7]8
[4]8
[1]8
[6]8
[3]8
[6]8
[0]8
[6]8
[4]8
[2]8
[0]8
[6]8
[4]8
[2]8
[7]8
[0]8
[7]8
[6]8
[5]8
[4]8
[3]8
[2]8
[1]8
For example, the row for [4]8 does not contain every element.
As discussed above, if we “throw away” the non-invertible elements we
get a group, (Z∗8 , ×) ,
×
[1]8
[3]8
[5]8
[7]8
[1]8
[1]8
[3]8
[5]8
[7]8
[3]8
[3]8
[1]8
[7]8
[5]8
[5]8
[5]8
[7]8
[1]8
[3]8
[7]8
[7]8
[5]8
[3]8
[1]8
Unfortunately there is no way to look at the table of a closed binary
operation on a finite set to see if the operation is associative.
Finally, a group is abelian if, and only if, the table is symmetric about
its leading diagonal.
10
Appendix
Contents
• Elementary consequences of the Group axioms:
• Maps between Groups
• Rings
• Fields
• More examples of binary operations and groups.
Elementary consequences of the Group axioms:
Theorem Let (G, ∗) be a group.
(i) For a, x, y ∈ G if x ∗ a = y ∗ a then x = y (a cancellation result),
(ii) For a, x, y ∈ G if a ∗ x = a ∗ y then x = y (a cancellation result),
(iii) e−1 = e, where e is the identity,
−1
(iv) For all x ∈ G, (x−1 ) = x,
(v) For all x, y ∈ G, (x ∗ y)−1 = y −1 ∗ x−1 ,
(vi) For x1 , x2 , ..., xn ∈ G,
−1
−1
(x1 ∗ x2 ∗ ... ∗ xn )−1 = x−1
n ∗ ... ∗ x2 ∗ x1 ,
(vii) For m, n ∈ Z and g ∈ G, g m ∗ g n = g m+n and (g n )m = g nm .
Proof (i)
⇒
⇒
⇒
⇒
x∗a=y∗a
(x ∗ a) ∗ a−1 = (y ∗ a) ∗ a−1
x ∗ a ∗ a−1 = y ∗ a ∗ a−1 by associativity,
x ∗ e = y ∗ e definition of inverses,
x = y.
⇒
⇒
⇒
⇒
a∗x=a∗y
a−1 ∗ (a ∗ x) = a−1 ∗ (a ∗ y)
a−1 ∗ a ∗ x = a−1 ∗ a ∗ y by associativity,
e ∗ x = e ∗ y definition of inverses,
x = y.
(ii)
11
(iii)
e−1 = e−1 ∗ e since e is the identity,
= e since e−1 is the inverse of e.
−−1
(iv) By definition (x−1 )
is the inverse of x−1 . Yet x−1 ∗ x = x ∗ x−1 = e
means that x is also the inverse of x−1 . We know that the inverse of an
−1
element is unique, hence (x−1 ) = x as required.
(v) By definition (x ∗ y)−1 is the inverse of x ∗ y. Yet
(x ∗ y) ∗ y −1 ∗ x−1 = (x ∗ y) ∗ y −1 ∗ x−1 by associativity,
= x ∗ y ∗ y −1 ∗ x−1 again by associativity,
= (x ∗ e) ∗ x−1
= x ∗ x−1 = e.
Similarly, (y −1 ∗ x−1 ) ∗ (x ∗ y) = e. So y −1 ∗ x−1 is also an inverse of x ∗ y. We
know that the inverse of an element is unique, hence (x ∗ y)−1 = y −1 ∗ x−1 as
required.
(vi) Use induction based on
(x1 ∗ x2 ∗ ... ∗ xn )−1 = ((x1 ∗ x2 ∗ ... ∗ xn−1 ) ∗ xn )−1
−1
= x−1
,
n ∗ (x1 ∗ ... ∗ xn−1 )
having used part (v).
(vi) Proof of g m ∗ g n = g m+n : If m ≥ 1, n ≥ 1 the result has been seen
earlier in the Lemma for semigroups.
If either m = 0 or n = 0, use the fact that g 0 = e, the identity.
If m ≤ −1, n ≤ −1 write m = −r, n = −t when, by the definition for
negative powers,
r
s
g m ∗ g n = g −1 ∗ g −1
r+s
= g −1
by the result for semigroups,
= g −(r+s) = g m+n .
If either m ≤ −1 and n ≥ 1 or m ≥ 1 and n ≤ −1 we have g m ∗ g n equalling
12
r
s
either (g −1 ) ∗ g n or g m ∗ (g −1 ) . Consider
r−1 −1 r
∗g
∗ g ∗ g n−1
g −1
g −1 ∗ g n =
by definition of n − th power and by (3) above,
r−1 −1 n−1
∗g
∗g ∗g
=
g −1
associativity,
r−1
∗ g −1 ∗ g ∗ g n−1
=
g −1
associativity
r−1 n−1
=
g −1
∗e ∗g
r−1 n−1
∗g .
= g −1
r−n
Continue, to get (g −1 ) , if r ≥ n, or g n−r otherwise. In both cases we get
s
g n−r = g n+m since m = −r. The similar result follows from g m ∗ (g −1 ) ,
namely it equals g m−s = g m+n since n = −s.
In all cases we have g m ∗ g n = g m+n .
Proof of (g n )m = g mn : If m ≥ 1, n ≥ 1 the result has been seen earlier
in the Lemma for semigroups.
If either m = 0 or n = 0, both sides are equal to the identity.
If n ≥ 1 and m ≤ −1 write m = −r. Then
r
(g n )m = (g n )−r = (g n )−1 by definition of negative exponent,
n r
=
g −1
, by part (vi) of this Theorem,
−1 nr
= g
by this result for positive exponents,
−nr
= g
by definition of negative exponent,
nm
= g .
If n ≤ −1 and m ≥ 1 write n = −s. Then
m
s m
(g n )m = g −s = g −1
by definition of negative exponent,
−1 sm
= g
by this result for positive exponents,
−sm
= g
by definition of negative exponent,
nm
= g .
13
If n ≤ −1 and m ≤ −1 then
−r −1 s −1 r
(g n )m = g −s
=
g
by definition of negative exponent,
−1 r
=
(g s )−1
by part (vi) of this Theorem,
= (g s )r by part (iv) of this Theorem,
= g sr by this result for positive exponents,
= g mn since mn = rs.
Hence in all cases (g n )m = g mn ..
Maps between groups.
At the end of Chapter 4 there was a vague comment that the multiplication tables for (Z∗8 , ×) and (Z∗5 , ×) are in some way different. We then ask,
what of the tables for (Z∗5 , ×) and (Z4 , +), written as
×
[1]5
[2]5
[3]5
[4]5
[1]5
[1]5
[2]5
[3]5
[4]5
[2]5
[2]5
[4]5
[1]5
[3]5
[3]5
[3]5
[1]5
[4]5
[2]5
[4]5
+
[4]5
[0]4
[3]5 and [1]4
[3]4
[2]5
[1]5
[2]4
[0]4
[0]4
[1]4
[3]4
[2]4
[1]4
[1]4
[2]4
[0]4
[3]4
[3]4
[3]4
[0]4
[2]4
[1]4
[2]4
[2]4
[3]4 .
[1]4
[0]4
(4)
Do they not have the same “form”? They do because there is a bijection θ
between the tables given by
θ ([1]5 ) = [0]4 , θ ([2]5 ) = [1]4 , θ ([3]5 ) = [3]4 and θ ([4]5 ) = [2]4 .
This has the property that θ ([a]5 × [b]5 ) = θ ([a]5 ) + θ ([b]5 ) , for all a, b ∈
{1, 2, 3, 4} . So not only are the sets mapped to each other, but the tables are
mapped to each other.
In general, let (G, ∗) and (H, ◦) be two groups (so, may be different sets
and different binary operations). If they are the “same” in the way above we
have a bijection θ : G → H which maps the multiplication table for G onto
that for H. So
(G, ∗)
..
.
..
.
g1
..
.
g2
..
..
.
.
···
..
.
g1 ∗ g2
..
.
···
···
···
(H, ◦)
..
.
θ
..
→
.
h1 = θ (g1 )
..
.
14
h2 = θ (g2 ) · · ·
..
..
.
.
···
..
.
h1 ◦ h2
..
.
···
,
···
thus g1 ∗ g2 must map onto h1 ∗ h2 . This means that
θ (g1 ∗ g2 ) = h1 ◦ h2 = θ (g1 ) ◦ θ (g2 ) .
Something for future years: Study maps θ : G → H satisfying θ (g1 ∗ g2 ) =
θ (g1 ) ◦ θ (g2 ), for all g1 , g2 in a group. Such maps are called isomorphisms,
and if there exists an isomorphism between two groups, they are called isomorphic.
It can be hard to find isomorphisms. Note that in (4) we wrote the table
for (Z4 , +) in the non-standard order of [0]4 , [1]4 , [3]4 and [2]4 to see the
similarity with (Z∗5 , ×). Imagine how hard it might be to see this similarity
if the tables are far larger, or the groups are infinite!
Sometimes it can be easier to show that two groups are not isomorphic.
It can be shown that if (G, ∗) and (H, ◦) are isomorphic with isomorphism θ
then if h = θ (g), h and g have the same orders. So the orders of elements in
(G, ∗) must match the orders of elements in (H, ◦) .
Example In (Z∗8 , ×) every element, other than [1]8 has order 2. Above we
saw that in (Z∗5 , ×), the element [2]5 has order 4. Hence these groups are not
isomorphic, a result alluded to at the end of Chapter 4.
Further study would be to find how many non-isomorphic groups there
are of each order. The first few results are
n=6
n = 8,
n = 9,
n = 10,
n = 12,
2
5
2
2
5
groups,
groups,
groups,
groups,
groups.
Rings
Definition A ring is a non-empty set R along with two binary operations
on R, addition + and multiplication ×, such that
(i) (R, +) is an abelian group,
(ii) R is closed under multiplication,
(iii) multiplication is associative on R
(iv) For all a, b, c ∈ R we have
a × (b + c) = a × b + a × c
(b + c) × a = b × a + c × a.
15
These are called the Distributive laws, we are “distributing” the a throughout the terms of the bracket. These laws are important in that they combine
both operations, + and −.
Note that we don’t demand that (R, ×) is a group. Non-zero elements in
R may fail to have inverses. Even more basic, there may not be a multiplicative identity! And we also don’t demand that multiplication is commutative.
Examples of rings: (i) (Z, +, ×) is the first example of a ring. From that
we can go to a polynomial ring (Z [x] , +, ×) and a finite ring (Zm , +, ×) .
(ii) The set of n × n matrices with real coefficients, (Mn (R) , +, ×) , is a
ring. We have a multiplicative identity, the identity matrix, but not every
matrix has an inverse. Also this ring is non-commutative.
The interest in rings lies in how the two operations interact with each
other. We saw before, in the section on primes, that if we ask additive
questions about multiplicative objects, primes, we get questions, such as
Goldbach’s Conjecture, that have withstood hundreds of years.
In the case of a general ring we have the additive identity 0. What
would we expect of the multiplication 0 × a for a ∈ R? Unsurprisingly
Example For all a ∈ R we have 0 × a = 0.
Solution Let a ∈ R be given. Since 0 is the additive identity, we have
0 + 0 = 0. Then
(0 + 0) × a = 0 × a
0 × a + 0 × a = 0 × a, by distributive law,
0 × a = 0, on adding − (0 × a) to both sides.
Since the distributive law is the only axiom to contain both operations,
+ and × it is no surprise we have to use it in the proof above.
Again, −1 is the additive inverse of 1 as is −a for any element of R. But
are −a and −1 × a the same? Unsurprisingly
Example For all a ∈ R we have −1 × a = −a.
Solution Let a ∈ R be given. Start from 1 + (−1) = 0. Then
(1 + (−1)) × a
1 × a + (−1) × a
a + (−1) × a
−a + (a + (−1) × a)
(−a + a) + (−1) × a
0 + (−1) × a
(−1) × a
=
=
=
=
=
=
=
0 × a = 0 by result above,
0 by distributive law,
0 since 1 is the multiplicative identity,
−a + 0 adding − a to both side,
−a, by associativity on LHS and 0 identity on RHS,
−a,
−a,
16
What we can see here is that because we have only a few axioms defining
a ring, the proof of something quite familiar, is surprisingly long. We increase
the number of axioms in the next section.
Fields
Definition A field (F, +, ×) , is a non-empty set F along with two binary
operations on F , addition + and multiplication ×, such that satisfies
(i) (F, +, ×) is a ring
(ii) multiplication is commutative on F ,
(iii) There is a multiplicative identity in F,
(iv) Every non-zero element of F has a multiplicative inverse.
The interest in fields comes from the fact that many of a our familiar
arithmetic structures are fields.
Examples of fields:
(i) (R, +, ×) , (C, +, ×) and (Q, +, ×) are infinite fields,
(ii) (Zp , +p , ×p ) for p a prime, is an example of a finite field.
But note that (Zm , +m , ×m ) is not a field if m not prime.
Something for future years: (R, +, ×) , (C, +, ×) and (Q, +, ×) satisfy the
axioms of a field and so, to this extent, are the same. But what further
properties do they satisfy that show they are different?
On (R, +, ×) and (Q, +, ×) we can define an order relation a < b with
properties such that if a < b then a + c < b + c and if a < b and b < c then
a < c. It can be shown that no such relation exists on (C, +, ×). Hence
(C, +, ×) is different to both (R, +, ×) and (Q, +, ×).
But are (R, +, ×) and (Q, +, ×) different? From the first half of the course
you know they are different, R is uncountable while Q is countable. But here
I want to mention another difference. Consider the sequence of rational
numbers 1, 1.4, 1.41, 1.414,
√ 1.4142, 1.41421, ....√, which gets arbitrarily close
/ Q. So, in (Q, +, ×) it is
(converges to) the limit 2. Unfortunately, 2 ∈
not true that all convergent sequences converge. But it can be shown that in
(R, +, ×) all convergent sequences converge. These ideas take us away from
algebra and into the areas of analysis.
More examples of binary operations and groups.
Example of binary operations not seen in course:
S = Mn (R), the set of n×n matrices with real entries, with either matrix
addition or matrix multiplication.
17
Let X be any non-empty set, S = P (X) , the power set of X, with either
∩ or ∪.
Let Ω be a non-empty set and S the set of all functions Ω → Ω along
with ◦, composition of functions.
Example Not a binary operation:
If S = N then subtraction − is not a binary operation since 1 − 2 ∈
/ N.
a
If S = Q then a ∗ b = b is not a binary operation since 1 ∗ 0 has no
meaning.
Example of binary operations not seen in course: Want to take the time to
define sets of polynomials.
Definition For a set F , a polynomial over F with variable x is of the
form
an xn + an−1 xn−1 + an−2 xn−2 + ... + a1 x + a0 ,
where an , an−1 , ..., a1 , a0 ∈ F . The ai , 0 ≤ i ≤ n are the coefficients of
the polynomial. If xn is the largest power of x appearing in the polynomial
then n is the degree of the polynomial, an xn is the leading term and an is
the leading coefficient. The collection of all polynomials with one variable
x and with coefficients from F will be denoted by F [x] . (Note the square
brackets.)
Note that 0 ∈ F [x], being ... + 0x2 + 0x + 0, but it is not said to have a
degree, though some books give it degree −1 or even −∞.
Examples
3x2 + 5x − 1 ∈ Z [x] , x2 − π ∈ R [x] ,
5 2
3 3
x − 12
x + x ∈ Q [x] , 5x4 + x + 2 ∈ Z7 [x] .
7
If we can add and multiply numbers in the set F then we can add and
multiply the polynomials in F [x].
Examples (i) In Z [x] the sum of 3x2 + 5x − 1 and 5x3 − 3x2 + 2x + 1 is
3x2 + 5x − 1
+ 5x3 − 3x2 + 2x + 1
= 5x3
+ 7x.
(ii) Addition in Z3 [x]. The sum of 2x3 + 2x2 + x + 1 and x3 + 2x2 + 2 is
2x3 + 2x2 + x + 1
+ x3 + 2x2
+2
=
x2 + x,
18
(iii) In Z [x] the product of x2 + 2x + 3 and x2 + 4 is
x2 + 2x + 3 x2 + 4 =
3x2
+ 12
3
+ 2x
+ 8x
4
2
+x
+ 4x
4
3
= x + 2x + 7x2 + 8x + 12
(iv) Multiplication in Z2 [x] . The product of x3 + x + 1 and x2 + x + 1 is
x3 + x + 1 x2 + x + 1 = x5 + x4 + x3
+ x3 + x2 + x
+ x2 + x + 1
= x5 + x4
+ 1.
using 2 ≡ 0 mod 2. Thus + and × are binary operations on Z [x] and Zm [x].
Aside The results of Chapters 1-3, on arithmetic, congruencies and congruence classes can be given for either Z [x], Zm [x] in place of Z. This is
because we can talk of one polynomial dividing another.
Definition If f, g ∈ F [x], we say that g divides f if there exists h ∈ F [x]
such that f = gh.
Example In Z [x] , x − 1 divides x3 − 2x2 + 1 since
x3 − 2x2 + 1 = (x − 1) x2 − x − 1 .
In Z2 [x] , x + 1 divides x3 + 1 since
x3 + 1 = (x + 1) x2 + x + 1 .
We could then talk of greatest common divisors (greatest in terms of
degree) and linear combinations. Or we could talk about congruencies, saying
f (x) ≡ g (x) mod h (x) iff h (x) divides f (x) − g (x) and congruence classes.
We could then construct new “algebraic structures” by defining addition and
multiplication on these congruence classes of polynomials. This is something
for future years.
End of aside
Example of identity. In (P (X) , ∩) the identity is X since X ∩C = C ∩X =
C for all C ∈ P (X) .
Question Do we always have identities?
19
Examples Let 2Z be the set of even integers. The product of two even
integers is even so × is a binary operation on 2Z. Yet there is no identity in
(2Z, ×) (because 1 is not even).
In (Z, −) we have a right identity, n − 0 = n, but no left identity (the left
identity won’t be 0 since 0 − n = −n 6= n when n 6= 0 and no other possible
value for the left identity will work).
Question Do we always have inverses?
Example In (M2 (R) , ×) not every non-zero matrix here has an inverse, for
example
1 1
.
1 1
Examples of semigroups. (X is a non-empty set)
(Z, ×) , (Zn , +) , (Zn , ×) (N, +) , (2Z, ×) , (P (X) , ∩) , (P (X) , ∪) , (Sn , ◦) ,
(Z [x] , +) , (Z [x] , ×) , (Zn [x] , +) and (Zn [x] , ×) .
Definition An important subset of (M2 (R) , ×) is the collection of matrices
that have an inverse, i.e. are invertible. Such matrices have a non-zero
determinant.
a b
GL2 (R) =
: ad − bc 6= 0 .
c d
Here GL stands for General Linear. So again we have “thrown away”
the elements with no inverse.
Example In (M2 (R) , ×) let
1 1
1 0
a=
,b =
.
0 1
1 1
Then
2
(ab) =
2 1
1 1
1 0
2 1
2
But
2 2
ab =
1 2
0 1
=
5 3
3 2
=
.
5 2
2 1
.
So we don’t necessarily have a2 ∗ b2 = (a ∗ b)2 .
Example (Z, +) is an additive group.
Verification If m, n ∈ Z then m + n ∈ Z and so G1 holds.
If m, n, p ∈ Z then (m + n) + p = m + (n + p) and so G2 holds.
20
We have 0 ∈ Z and for all n ∈ Z, n + 0 = 0 + n = n and so G3 holds.
For any n ∈ Z we have −n ∈ Z and n + (−n) = (−n) + n = 0 and so G4
holds.
Thus (Z, +) is an additive group with identity 0 and the inverse of n is
−n.
Examples: (Zm , +m ) is an additive group.
Verification We know (Zm , +) is a semigroup. So only need note that [0]m
is the identity and, for [a]m ∈ Zm , the inverse is [−a]m = [m − a]m .
Example (Mn (R) , +) (Z [x] , +) and (Zm [x] , +) are further examples of
additive groups. In such groups the identity is normally denoted by 0.
Example ({i, −1, −i, 1} , ×), where i2 = −1, is a multiplicative group.
Verification From the table
1
i
×
1
1
i
i −1
i
−i −i 1
−1 −1 −i
−i −1
−i −1
1 −i
−1 i
i
1
we see that G1, G3 (with e = 1) and G4 (with 1−1 = 1, i−1 = −i, (−1)−1 =
−1 and (−i)−1 = i) are all satisfied. G2 holds since multiplication of complex
numbers is associative.
Examples
Other multiplicative groups are (C \ {0} , ×) , (Q \ {0} , ×) , and
∗
Zp , ×p .
Question for students. Why are (Mn (R) , ×) (Z [x] , ×) and (Zm [x] , ×) not
multiplicative groups?
Note In the theory of groups a general group is normally written with a
multiplicative operation.
In multiplicative groups the identity is normally denoted by 1, id or I.
Note that I say “normally”.
Example ({2, 4, 6, 8} , ×10 ) .
×10
2
4
6
8
2
4
8
2
6
4
8
6
4
2
6
2
4
6
8
8
6
2
8
4
From the table we see that the identity is 6. Also 2−1 = 8, 4−1 = 4, 6−1 = 6
and 8−1 = 2.
21