2.1
Groups, Rings, and Fields
In the following three chapters, the basic algebraic structures (groups, rings, fields, vector
spaces) are reviewed, with a major emphasis on vector spaces. Basic notions of linear algebra
such as vector spaces, subspaces, linear combinations, linear independence, bases, quotient
spaces, linear maps, matrices, change of bases, direct sums, linear forms, dual spaces, hyper-
planes, transpose of a linear maps, are reviewed.
The set R of real numbers has two operations + : R × R → R (addition) and ∗: R × R →
R (multiplication) satisfying properties that make R into an abelian group under +, and
∗
R − {0} = R into an abelian group under ∗. Recall the definition of a group.
Definition 2.1. A group is a set G equipped with a binary operation · : G × G → G that
associates an element a · b ∈ G to every pair of elements a, b ∈ G, and having the following
properties: · is associative, has an identity element e ∈ G, and every element in G is invertible
(w.r.t. ·). More explicitly, this means that the following equations hold for all a, b, c ∈ G:
(G1) a · (b · c) = (a · b) · c.
(associativity);
(G2) a · e = e · a = a.
(identity);
(G3) For every a ∈ G, there is some a−1 ∈ G such that a · a−1 = a−1 · a = e
(inverse).
A group G is abelian (or commutative) if
a · b = b · a
for all a, b ∈ G.
A set M together with an operation ·: M × M → M and an element e satisfying only
conditions (G1) and (G2) is called a monoid . For example, the set N = {0, 1, . . . , n, . . .} of
natural numbers is a (commutative) monoid under addition. However, it is not a group.
Some examples of groups are given below.
11
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
Example 2.1.
1. The set Z = {. . . , −n, . . . , −1, 0, 1, . . . , n, . . .} of integers is a group under addition,
with identity element 0. However, ∗
Z = Z − {0} is not a group under multiplication.
2. The set Q of rational numbers (fractions p/q with p, q ∈ Z and q = 0) is a group
under addition, with identity element 0. The set
∗
Q = Q − {0} is also a group under
multiplication, with identity element 1.
3. Similarly, the sets R of real numbers and C of complex numbers are groups under
addition (with identity element 0), and
∗
∗
R = R − {0} and C = C − {0} are groups
under multiplication (with identity element 1).
4. The sets
n
n
R and C of n-tuples of real or complex numbers are groups under compo-
nentwise addition:
(x1, . . . , xn) + (y1, . . . , yn) = (x1 + yn, . . . , xn + yn),
with identity element (0, . . . , 0). All these groups are abelian.
5. Given any nonempty set S, the set of bijections f : S → S, also called permutations
of S, is a group under function composition (i.e., the multiplication of f and g is the
composition g ◦ f), with identity element the identity function idS. This group is not
abelian as soon as S has more than two elements.
6. The set of n × n matrices with real (or complex) coefficients is a group under addition
of matrices, with identity element the null matrix. It is denoted by Mn(R) (or Mn(C)).
7. The set R[X] of polynomials in one variable with real coefficients is a group under
addition of polynomials.
8. The set of n × n invertible matrices with real (or complex) coefficients is a group under
matrix multiplication, with identity element the identity matrix In. This group is
called the general linear group and is usually denoted by GL(n, R) (or GL(n, C)).
9. The set of n ×n invertible matrices with real (or complex) coefficients and determinant
+1 is a group under matrix multiplication, with identity element the identity matrix
In. This group is called the special linear group and is usually denoted by SL(n, R)
(or SL(n, C)).
10. The set of n × n invertible matrices with real coefficients such that RR = In and of
determinant +1 is a group called the special orthogonal group and is usually denoted
by SO(n) (where R
is the transpose of the matrix R, i.e., the rows of R
are the
columns of R). It corresponds to the rotations in
n
R .
2.1. GROUPS, RINGS, AND FIELDS
13
11. Given an open interval ]a, b[, the set C(]a, b[) of continuous functions f : ]a, b[→ R is a
group under the operation f + g defined such that
(f + g)(x) = f (x) + g(x)
for all x ∈]a, b[.
It is customary to denote the operation of an abelian group G by +, in which case the
inverse a−1 of an element a ∈ G is denoted by −a.
The identity element of a group is unique. In fact, we can prove a more general Fact:
Fact 1. If a binary operation · : M × M → M is associative and if e ∈ M is a left identity
and e ∈ M is a right identity, which means that
e · a = a for all a ∈ M
(G2l)
and
a · e = a for all a ∈ M,
(G2r)
then e = e .
Proof. If we let a = e in equation (G2l), we get
e · e = e ,
and if we let a = e in equation (G2r), we get
e · e = e ,
and thus
e = e · e = e ,
as claimed.
Fact 1 implies that the identity element of a monoid is unique, and since every group is
a monoid, the identity element of a group is unique. Furthermore, every element in a group
has a unique inverse. This is a consequence of a slightly more general fact:
Fact 2. In a monoid M with identity element e, if some element a ∈ M has some left inverse
a ∈ M and some right inverse a ∈ M, which means that
a · a = e
(G3l)
and
a · a = e,
(G3r)
then a = a .
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
Proof. Using (G3l) and the fact that e is an identity element, we have
(a · a) · a = e · a = a .
Similarly, Using (G3r) and the fact that e is an identity element, we have
a · (a · a ) = a · e = a .
However, since M is monoid, the operation · is associative, so
a = a · (a · a ) = (a · a) · a = a ,
as claimed.
Remark: Axioms (G2) and (G3) can be weakened a bit by requiring only (G2r) (the exis-
tence of a right identity) and (G3r) (the existence of a right inverse for every element) (or
(G2l) and (G3l)). It is a good exercise to prove that the group axioms (G2) and (G3) follow
from (G2r) and (G3r).
If a group G has a finite number n of elements, we say that G is a group of order n. If
G is infinite, we say that G has infinite order . The order of a group is usually denoted by
|G| (if G is finite).
Given a group, G, for any two subsets R, S ⊆ G, we let
RS = {r · s | r ∈ R, s ∈ S}.
In particular, for any g ∈ G, if R = {g}, we write
gS = {g · s | s ∈ S}
and similarly, if S = {g}, we write
Rg = {r · g | r ∈ R}.
From now on, we will drop the multiplication sign and write g1g2 for g1 · g2.
For any g ∈ G, define Lg, the left translation by g, by Lg(a) = ga, for all a ∈ G, and
Rg, the right translation by g, by Rg(a) = ag, for all a ∈ G. Observe that Lg and Rg are
bijections. We show this for Lg, the proof for Rg being similar.
If Lg(a) = Lg(b), then ga = gb, and multiplying on the left by g−1, we get a = b, so Lg
injective. For any b ∈ G, we have Lg(g−1b) = gg−1b = b, so Lg is surjective. Therefore, Lg
is bijective.
Definition 2.2. Given a group G, a subset H of G is a subgroup of G iff
(1) The identity element, e, of G also belongs to H (e ∈ H);
2.1. GROUPS, RINGS, AND FIELDS
15
(2) For all h1, h2 ∈ H, we have h1h2 ∈ H;
(3) For all h ∈ H, we have h−1 ∈ H.
The proof of the following proposition is left as an exercise.
Proposition 2.1. Given a group G, a subset H ⊆ G is a subgroup of G iff H is nonempty
and whenever h1, h2 ∈ H, then h1h−1
2
∈ H.
If the group G is finite, then the following criterion can be used.
Proposition 2.2. Given a finite group G, a subset, H ⊆ G is a subgroup of G iff
(1) e ∈ H;
(2) H is closed under multiplication.
Proof. We just have to prove that condition (3) of Definition 2.2 holds. For any a ∈ H, since
the left translation La is bijective, its restriction to H is injective, and since H is finite, it is
also bijective. Since e ∈ H, there is a unique b ∈ H such that La(b) = ab = e. However, if
a−1 is the inverse of a in G, we also have La(a−1) = aa−1 = e, and by injectivity of La, we
have a−1 = b ∈ H.
Definition 2.3. If H is a subgroup of G and g ∈ G is any element, the sets of the form gH
are called left cosets of H in G and the sets of the form Hg are called right cosets of H in
G.
The left cosets (resp. right cosets) of H induce an equivalence relation, ∼, defined as
follows: For all g1, g2 ∈ G,
g1 ∼ g2 iff g1H = g2H
(resp. g1 ∼ g2 iff Hg1 = Hg2). Obviously, ∼ is an equivalence relation.
Now, we claim that g1H = g2H iff g−1
2 g1H = H iff g−1
2 g1 ∈ H .
If we apply the bijection L
to both g
(g
g−1
1H and g2H we get Lg−1
1H ) = g−1
2 g1H and
2
2
L
(g
g−1
2H ) = H , so g1H = g2H iff g−1
2 g1H = H .
If g−1
2 g1H = H , since 1 ∈ H , we get
2
g−1
2 g1 ∈ H . Conversely, if g−1
2 g1 ∈ H , since H is a group, the left translation L
is a
g−1g
2
1
bijection of H, so g−1
2 g1H = H . Thus, g−1
2 g1H = H iff g−1
2 g1 ∈ H .
It follows that the equivalence class of an element g ∈ G is the coset gH (resp. Hg).
Since Lg is a bijection between H and gH, the cosets gH all have the same cardinality. The
map Lg−1 ◦ Rg is a bijection between the left coset gH and the right coset Hg, so they also
have the same cardinality. Since the distinct cosets gH form a partition of G, we obtain the
following fact:
Proposition 2.3. (Lagrange) For any finite group G and any subgroup H of G, the order
h of H divides the order n of G.
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
The ratio n/h is denoted by (G : H) and is called the index of H in G. The index (G : H)
is the number of left (and right) cosets of H in G. Proposition 2.3 can be stated as
|G| = (G : H)|H|.
The set of left cosets of H in G (which, in general, is not a group) is denoted G/H.
The “points” of G/H are obtained by “collapsing” all the elements in a coset into a single
element.
It is tempting to define a multiplication operation on left cosets (or right cosets) by
setting
(g1H)(g2H) = (g1g2)H,
but this operation is not well defined in general, unless the subgroup H possesses a special
property. This property is typical of the kernels of group homomorphisms, so we are led to
Definition 2.4. Given any two groups, G, G , a function ϕ : G → G is a homomorphism iff
ϕ(g1g2) = ϕ(g1)ϕ(g2),
for all g1, g2 ∈ G.
Taking g1 = g2 = e (in G), we see that
ϕ(e) = e ,
and taking g1 = g and g2 = g−1, we see that
ϕ(g−1) = ϕ(g)−1.
If ϕ : G → G and ψ : G → G are group homomorphisms, then ψ ◦ ϕ: G → G is also a
homomorphism. If ϕ : G → G is a homomorphism of groups and H ⊆ G and H ⊆ G are
two subgroups, then it is easily checked that
Im H = ϕ(H) = {ϕ(g) | g ∈ H} is a subgroup of G
(Im H is called the image of H by ϕ) and
ϕ−1(H ) = {g ∈ G | ϕ(g) ∈ H } is a subgroup of G.
In particular, when H = {e }, we obtain the kernel, Ker ϕ, of ϕ. Thus,
Ker ϕ = {g ∈ G | ϕ(g) = e }.
It is immediately verified that ϕ : G → G is injective iff Ker ϕ = {e}. (We also write
Ker ϕ = (0).) We say that ϕ is an isomorphism if there is a homomorphism, ψ : G → G, so
that
ψ ◦ ϕ = idG and ϕ ◦ ψ = idG .
2.1. GROUPS, RINGS, AND FIELDS
17
In this case, ψ is unique and it is denoted ϕ−1. When ϕ is an isomorphism we say the
the groups G and G are isomorphic. It is easy to see that a bijective hmomorphism is an
isomorphism. When G = G, a group isomorphism is called an automorphism.
The left translations Lg and the right translations Rg are group isomorphisms.
We claim that H = Ker ϕ satisfies the following property:
gH = Hg,
for all g ∈ G.
(∗)
First, note that (∗) is equivalent to
gHg−1 = H,
for all g ∈ G,
and the above is equivalent to
gHg−1 ⊆ H, for all g ∈ G.
(∗∗)
This is because gHg−1 ⊆ H implies H ⊆ g−1Hg, and this for all g ∈ G. But,
ϕ(ghg−1) = ϕ(g)ϕ(h)ϕ(g−1) = ϕ(g)e ϕ(g)−1 = ϕ(g)ϕ(g)−1 = e ,
for all h ∈ H = Ker ϕ and all g ∈ G. Thus, by definition of H = Ker ϕ, we have gHg−1 ⊆ H.
Definition 2.5. For any group, G, a subgroup, N ⊆ G, is a normal subgroup of G iff
gN g−1 = N,
for all g ∈ G.
This is denoted by N
G.
Observe that if G is abelian, then every subgroup of G is normal.
If N is a normal subgroup of G, the equivalence relation induced by left cosets is the
same as the equivalence induced by right cosets. Furthermore, this equivalence relation, ∼,
is a congruence, which means that: For all g1, g2, g1, g2 ∈ G,
(1) If g1N = g1N and g2N = g2N, then g1g2N = g1g2N, and
(2) If g1N = g2N, then g−1
1 N = g−1
2 N .
As a consequence, we can define a group structure on the set G/ ∼ of equivalence classes
modulo ∼, by setting
(g1N)(g2N) = (g1g2)N.
This group is denoted G/N and called the quotient of G by N . The equivalence class, gN ,
of an element g ∈ G is also denoted g (or [g]). The map π : G → G/N given by
π(g) = g = gN,
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
is clearly a group homomorphism called the canonical projection.
Given a homomorphism of groups, ϕ : G → G , we easily check that the groups G/Ker ϕ
and Im ϕ = ϕ(G) are isomorphic. This is often called the first isomorphism theorem.
A useful way to construct groups is the direct product construction. Given two groups G
an H, we let G × H be the Cartestian product of the sets G and H with the multiplication
operation · given by
(g1, h1) · (g2, h2) = (g1g2, h1h2).
It is immediately verified that G × H is a group. Similarly, given any n groups G1, . . . , Gn,
we can define the direct product G1 × · · · × Gn is a similar way.
If G is an abelian group and H1, . . . , Hn are subgroups of G, the situation is simpler.
Consider the map
a : H1 × · · · × Hn → G
given by
a(h1, . . . , hn) = h1 + · · · + hn,
using + for the operation of the group G. It is easy to verify that a is a group homomorphism,
so its image is a subgroup of G denoted by H1 + · · · + Hn, and called the sum of the groups
Hi. The following proposition will be needed.
Proposition 2.4. Given an abelian group G, if H1 and H2 are any subgroups of G such
that H1 ∩ H2 = {0}, then the map a is an isomorphism
a : H1 × H2 → H1 + H2.
Proof. The map is surjective by definition, so we just have to check that it is injective. For
this, we show that Ker a = {(0, 0)}. We have a(a1, a2) = 0 iff a1 + a2 = 0 iff a1 = −a2. Since
a1 ∈ H1 and a2 ∈ H2, we see that a1, a2 ∈ H1 ∩ H2 = {0}, so a1 = a2 = 0, which proves that
Ker a = {(0, 0)}.
Under the conditions of Proposition 2.4, namely H1 ∩ H2 = {0}, the group H1 + H2 is
called the direct sum of H1 and H2; it is denoted by H1 ⊕ H2, and we have an isomorphism
H
∼
1 × H2 = H1 ⊕ H2.
The groups Z, Q, R, C, and Mn(R) are more than an abelian groups, they are also com-
mutative rings. Furthermore, Q, R, and C are fields. We now introduce rings and fields.
Definition 2.6. A ring is a set A equipped with two operations + : A × A → A (called
addition) and ∗ : A × A → A (called multiplication) having the following properties:
(R1) A is an abelian group w.r.t. +;
(R2) ∗ is associative and has an identity element 1 ∈ A;
(R3) ∗ is distributive w.r.t. +.
2.1. GROUPS, RINGS, AND FIELDS
19
The identity element for addition is denoted 0, and the additive inverse of a ∈ A is
denoted by −a. More explicitly, the axioms of a ring are the following equations which hold
for all a, b, c ∈ A:
a + (b + c) = (a + b) + c
(associativity of +)
(2.1)
a + b = b + a
(commutativity of +)
(2.2)
a + 0 = 0 + a = a
(zero)
(2.3)
a + (−a) = (−a) + a = 0
(additive inverse)
(2.4)
a ∗ (b ∗ c) = (a ∗ b) ∗ c
(associativity of ∗)
(2.5)
a ∗ 1 = 1 ∗ a = a
(identity for ∗)
(2.6)
(a + b) ∗ c = (a ∗ c) + (b ∗ c)
(distributivity)
(2.7)
a ∗ (b + c) = (a ∗ b) + (a ∗ c)
(distributivity)
(2.8)
The ring A is commutative if
a ∗ b = b ∗ a
for all a, b ∈ A.
From (2.7) and (2.8), we easily obtain
a ∗ 0 = 0 ∗ a = 0
(2.9)
a ∗ (−b) = (−a) ∗ b = −(a ∗ b).
(2.10)
Note that (2.9) implies that if 1 = 0, then a = 0 for all a ∈ A, and thus, A = {0}. The
ring A = {0} is called the trivial ring. A ring for which 1 = 0 is called nontrivial. The
multiplication a ∗ b of two elements a, b ∈ A is often denoted by ab.
Example 2.2.
1. The additive groups Z, Q, R, C, are commutative rings.
2. The group R[X] of polynomials in one variable with real coefficients is a ring under
multiplication of polynomials. It is a commutative ring.
3. The group of n × n matrices Mn(R) is a ring under matrix multiplication. However, it
is not a commutative ring.
4. The group C(]a, b[) of continuous functions f : ]a, b[→ R is a ring under the operation
f · g defined such that
(f · g)(x) = f(x)g(x)
for all x ∈]a, b[.
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
When ab = 0 with b = 0, we say that a is a zero divisor . A ring A is an integral domain
(or an entire ring) if 0 = 1, A is commutative, and ab = 0 implies that a = 0 or b = 0, for
all a, b ∈ A. In other words, an integral domain is a nontrivial commutative ring with no
zero divisors besides 0.
Example 2.3.
1. The rings Z, Q, R, C, are integral domains.
2. The ring R[X] of polynomials in one variable with real coefficients is an integral domain.
3.
4. For any positive integer, p ∈ N, define a relation on Z, denoted m ≡ n (mod p), as
follows:
m ≡ n (mod p) iff m − n = kp for some k ∈ Z.
The reader will easily check that this is an equivalence relation, and, moreover, it is
compatible with respect to addition and multiplication, which means that if m1 ≡ n1
(mod p) and m2 ≡ n2 (mod p), then m1 + m2 ≡ n1 + n2 (mod p) and m1m2 ≡ n1n2
(mod p). Consequently, we can define an addition operation and a multiplication
operation of the set of equivalence classes (mod p):
[m] + [n] = [m + n]
and
[m] · [n] = [mn].
Again, the reader will easily check that the ring axioms are satisfied, with [0] as zero
and [1] as multiplicative unit. The resulting ring is denoted by Z/pZ.1 Observe that
if p is composite, then this ring has zero-divisors. For example, if p = 4, then we have
2 · 2 ≡ 0 (mod 4).
However, the reader should prove that Z/pZ is an integral domain if p is prime (in
fact, it is a field).
5. The ring of n × n matrices Mn(R) is not an integral domain. It has zero divisors.
A homomorphism between rings is a mapping preserving addition and multiplication
(and 0 and 1).
1The notation Zp is sometimes used instead of Z/pZ but it clashes with the notation for the p-adic integers
so we prefer not to use it.
2.1. GROUPS, RINGS, AND FIELDS
21
Definition 2.7. Given two rings A and B, a homomorphism between A and B is a function
h : A → B satisfying the following conditions for all x, y ∈ A:
h(x + y) = h(x) + h(y)
h(xy) = h(x)h(y)
h(0) = 0
h(1) = 1.
Actually, because B is a group under addition, h(0) = 0 follows from
h(x + y) = h(x) + h(y).
Example 2.4.
1. If A is a ring, for any integer n ∈ Z, for any a ∈ A, we define n · a by
n · a = a + · · · + a
n
if n ≥ 0 (with 0 · a = 0) and
n · a = −(−n) · a
if n < 0. Then, the map h : Z → A given by
h(n) = n · 1A
is a ring homomorphism (where 1A is the multiplicative identity of A).
2. Given any real λ ∈ R, the evaluation map ηλ : R[X] → R defined by
ηλ(f (X)) = f (λ)
for every polynomial f (X) ∈ R[X] is a ring homomorphism.
A ring homomorphism h : A → B is an isomorphism iff there is a homomorphism g : B →
A such that g ◦ f = idA and f ◦ g = idB. Then, g is unique and denoted by h−1. It is easy
to show that a bijective ring homomorphism h : A → B is an isomorphism. An isomorphism
from a ring to itself is called an automorphism.
Given a ring A, a subset A of A is a subring of A if A is a subgroup of A (under
addition), is closed under multiplication, and contains 1. If h : A → B is a homomorphism
of rings, then for any subring A , the image h(A ) is a subring of B, and for any subring B
of B, the inverse image h−1(B ) is a subring of A.
A field is a commutative ring K for which A − {0} is a group under multiplication.
Definition 2.8. A set K is a field if it is a ring and the following properties hold:
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CHAPTER 2. VECTOR SPACES, BASES, LINEAR MAPS
(F1) 0 = 1;
(F2) K∗ = K − {0} is a group w.r.t. ∗ (i.e., every a = 0 has an inverse w.r.t. ∗);
(F3) ∗ is commutative.
If ∗ is not commutative but (F1) and (F2) hold, we say that we have a skew field (or
noncommutative field ).
Note that we are assuming that the operation ∗ of a field is commutative. This convention
is not universally adopted, but since ∗ will be commutative for most fields we will encounter,
we may as well include this condition in the definition.
Example 2.5.
1. The rings Q, R, and C are fields.
2. The set of (formal) fractions f (X)/g(X) of polynomials f (X), g(X) ∈ R[X], where
g(X) is not the null polynomial, is a field.
3. The ring C(]a, b[) of continuous functions f : ]a, b[→ R such that f(x) = 0 for all
x ∈]a, b[ is a field.
4. The ring Z/pZ is a field whenever p is prime.
A homomorphism h : K1 → K2 between two fields K1 and K2 is just a homomorphism
between the rings K1 and K2. However, because K∗1 and K∗2 are groups under multiplication,
a homomorphism of fields must be injective.
First, observe that for any x = 0,
1 = h(1) = h(xx−1) = h(x)h(x−1)
and
1 = h(1) = h(x−1x) = h(x−1)h(x),
so h(x) = 0 and
h(x−1) = h(x)−1.
But then, if h(x) = 0, we must have x = 0. Consequently, h is injective.
A field homomorphism h : K1 → K2 is an isomorphism