Algebraic Structures: Groups
Basic points on Group theory
Introduction Groups are the central objects of algebra. Even though the
definition of group is simple, it leads to a rich and amazing theory.Everything presented here is standard, except that the product ofgroups is given in the additive notation.
Definition
Suppose πΊ is a non-void set and Ο : πΊ x πΊ β πΊ is a function. Ο is called a binary operation, and we will write π (π, π) = π. π or π (π, π) = π + π. Consider the following properties.
1. If π, π, π β πΊ then π . (π . π) = (π . π) . π.
2. β π = πG β πΊ such that if π β πΊ: π . π = π . π = π.
3. If π β πΊ, βπ β πΊ with π . π = π . π = π, ( π = π-1).
4. If π, π β πΊ, then π . π = π . π.
Definition If properties 1), 2), and 3) hold, (πΊ, π) is said to be agroup. If we write π (π, π) = π . π, we say it is a multiplicativegroup. If we write π (π, π) = π + π, we say it is an additive group.If in addition, property 4) holds, we say the group is abelian orcommutative.
Groups
Theorem Let (πΊ, π) be a multiplicative group.
I. Suppose π, π, π β πΊ. Then π . π = π . π β π =
π . Also π . π = π . π β π = π.
II. π is unique, i.e., if π β πΊ satisfies 2), then π = π . Recall that b is an
identity in πΊ provided it is a right and left identity for any a in πΊ.
However, group structure is so rigid that if β π β πΊ such that π is a
right identity for π, then π = π. Of course, this is just a special case
of the cancellation law in (i).
III. Every right inverse is an inverse, i.e., if π . π = π then π = πβ1.
Also if π . π = π then π = πβ1. Thus inverses are unique.
IV. If π β πΊ, then (πβ1)β1 = π.
Groups
V. The multiplication π1 . π2 . π3 = π1 . (π2 . π3) = (π1 . π2) . π3 is
well-defined. In general, π1 . π2 . . . ππ is well defined.
VI. If π, π β πΊ, (ππ)β1= πβ1 . πβ1.
VII. Suppose a β G. Let π0 = π and if π > 0, ππ = π . . . π (π
times) and πβπ = πβ1β¦πβ1 (n times). If π1, π2, β¦ , ππ‘ β π then
ππ1ππ2 β¦πππ‘ = ππ1+π2+β―+ππ‘ . Also ππ π = πππ .
VIII. Finally, if G is abelian and a, b β G, then (ππ)π = ππ . ππ .
Exercises
Exercise. Write out the above theorem where πΊ is an additive group.
Note that part (vii) states that πΊ has a scalar multiplication over π.
This means that if a is in πΊ and π is an integer, there is defined an
element an in πΊ. This is so basic, that we state it explicitly.
Exercise. Suppose πΊ is a non-void set with a binary operation
π(π, π) = π . π which satisfies 1), 2) and [ 3') If π β πΊ, βπ β
πΊ with π . π = π]. Show (πΊ, π) is a group, i.e., show
π . π = π.
In other words, the group axioms are stronger than necessary. Ifevery element has a right inverse, then every element has a two sidedinverse.
Examples
Example πΊ = πΉ,πΊ = πΈ, or πΊ = π with π(π, π) = π + π is an additive abelian group.
Example πΊ = πΉβ 0 or πΊ = πΈβ 0 with π (π, π) = ππ is a multiplicative abelian group.
πΊ = πβ 0 with π (π, π) = ππ is not a group.
πΊ = {π β πΉ βΆ π > 0} with π (π, π) = ππ is a multiplicative abelian group.
Subgroups
Theorem Suppose πΊ is a multiplicative group and π» β πΊ is a non-void subset satisfying
1) If π, π β π» then π. π β π»
2) If π β π» then π-1β π»
Then π β π» and π» is a group under multiplication. H is called a subgroup of G.
Proof Since π» is non-void, βπ β π». By 2), π-1 β π» and so by 1), π βπ». The associative law is immediate and so π» is a group.
Examples
Example πΊ is a subgroup of πΊ and π is a subgroup of πΊ. These are called the improper subgroups of G.
Example If πΊ = π under addition, and π β π, then π» = ππ is a subgroup of π. By a theorem on the integers, every subgroup of π is of this form. This is a key property of the integers.
Exercises
Exercises Suppose πΊ is a multiplicative group.
a) Let π» be the center of πΊ, i.e., H = {β β πΊ βΆ π . β = β . π for all
π β πΊ}. Show π» is a subgroup of πΊ.
b) Suppose π»1 and π»2 are subgroups of πΊ. Show π»1 π»2 is a
subgroup of πΊ.
c) Suppose π»1 and π»2 are subgroups of πΊ, with neither π»1nor π»2
contained in the other. Show π»1 π π»2 is not a subgroup of πΊ.
Order
Suppose πΊ is a multiplicative group. If πΊ has an infinite number of elements, we say that π(πΊ), the order of πΊ, is infinite. If πΊ has n elements, then π(πΊ) = π. Suppose π β πΊ and π» = {ππ βΆ π βπ}.π» is an abelian subgroup of πΊ called the subgroup generated by a.
We define the order of the element π to be the order of H, i.e., the order of the subgroup generated by a. Let π βΆ π β π» be the surjective function defined by π (m) = am.
Note that π (k + π) = π π . π (π) where the addition is in Z and the multiplication is in the group H. We come now to the first real theorem in group theory. It says that the element a has finite order iff π is not injective, and in this case, the order of a is the smallest positive integer n with ππ = e.
Order
Theorem Suppose π is an element of a multiplicative group πΊ, andπ» = {ππ βΆ π β π}. If β distinct integers π and π with ππ = ππ,then a has some finite order n. In this case H has n distinctelements, H = {π0, π1,..., ππβ1}, and ππ = π iff π|π . In particular,the order of π is the smallest positive integer n with ππ = π, andπβ1(π) = ππ.
Proof Suppose π < π and ππ = ππ. Then ππβπ = π and thus β asmallest positive integer n with ππ = π. This implies that theelements of {π0, π1,..., ππβ1} are distinct, and we must show theyare all of H. If π β π, the Euclidean algorithm states that βintegers π and π with 0 β€ π < π and π = ππ + π . Thus ππ =πππ. ππ = ππ, and soπ» = {π0, π1,..., ππβ1}, and ππ = π iff π|π.
Exercises
Exercise Write out this theorem for πΊ an additive group. To begin, suppose a is an element of an additive group πΊ, and π» = { ππ βΆπ β π}.
Exercise Show that if πΊ is a finite group of even order, then πΊ has an odd number of elements of order 2. Note that e is the only element of order 1.
Cyclic group Definition A group πΊ is cyclic if β an element of G which generates
G.
Theorem If G is cyclic and H is a subgroup of G, then H is cyclic.
Cosets
Suppose H is a subgroup of a group G. It will be shown below that H
partitions G into right cosets. It also partitions G into left cosets, and
in general these partitions are distinct.
Theorem If H is a subgroup of a multiplicative group G, then
π ~ π defined by a ~ b iff π . πβ1 β π» is an equivalence relation.
If π β πΊ, ππ(π) = {π β πΊ βΆ π ~ π} = {β . π βΆ β β π»} =
π»π. Note thatπ . πβ1 β π» iff π . πβ1 β π».
Right and left cosets -Relations
Definition These equivalence classes are called right cosets. If the relation is defined by a ~ b iff b-1 . a β H, then the equivalence classes are cl(a) = aH and they are called left cosets. H is a left and right coset. If G is abelian, there is no distinction between right and left cosets. Note that b-1 . a β H iff a-1 . b β H.
In the theorem above, H is used to define an equivalence relation on G, and thus a partition of G. We now do the same thing a different way. We define the right cosets directly and show they form a partition of G.
Right and left cosets -Relations
Theorem Suppose H is a subgroup of a multiplicative group G. If a G G, define the right coset containing a to be Ha = {h . a : h β H}. Then the following hold.
1) Ha = π» iff a β H.
2) If π β π»π, then π»π = π»π, i.e., if β β π», then π»(β . π) =(Hh)a = Ha.
3) If π»π π»π β β , then π»π = π»π.
4) The right cosets form a partition of πΊ, i.e., each π in πΊ belongs to one and only one right coset.
5) Elements π and π belong to the same right coset iff π . π-1 β H iffπ . π-1 β H.
Index of the subgroup
Theorem Suppose H is a subgroup of a multiplicative group G.
a) Any two right cosets have the same number of elements. That is, if π, π β πΊ,
b) π βΆ π»π β π»π defined by π (β . π) = β . π is a bijection. Also any two left cosets have the same number of elements. Since H is a right and left coset, any two cosets have the same number of elements.
c) πΊ has the same number of right cosets as left cosets. The function F defined by πΉ(π»π) = π-1 π» is a bijection from the collection of right cosets to the left cosets. The number of right (or left) cosets is called the index of π» in G.
d) If πΊ is finite, π(π») (index of π») = π(πΊ) and so π(π») | π(πΊ). In other words, π(πΊ)/π(π») = the number of right cosets = the number of left cosets.
Index of the subgroup
If πΊ is finite, and π β πΊ, then π(π) | π(πΊ). (Proof: The order of a is
the order of the subgroup generated by π, and by 3) this divides the
order of πΊ.)
If πΊ has prime order, then πΊ is cyclic, and any element (except e) is a
generator. (Proof: Suppose π(πΊ) = π and π β πΊ, π β π. Then
π(π) | π and thus π(π) = π.)
If π(πΊ) = π and π β πΊ, then ππ = π. (Proof: ππ(π) = π and π =
π(π) (π(πΊ)/π(π)). )
Exercises
1. Suppose πΊ is a cyclic group of order 4, πΊ = {π, π, π2, π3} with π4 = π. Find the order of each element of πΊ. Find all the subgroups of πΊ.
2. Suppose πΊ is the additive group π and π» = 3π. Find the cosetsof π».
3. Think of a circle as the interval [0,1] with end points identified. Suppose πΊ = π under addition and π» = π. Show that the collection of all the cosets of π» can be thought of as a circle.
4. Let πΊ = π 2 under addition, and H be the subgroup defined by π» = {(π, 2π) βΆ π β π }. Find the cosets of π».
Normal Subgroups
We would like to make a group out of the collection of cosets of a
subgroup π». In general, there is no natural way to do that. However,
it is easy to do in case π» is a normal subgroup, which is described
below.
Theorem If π» is a subgroup of a group πΊ, then the following are
equivalent.
1. If π β πΊ, then ππ»πβ1 = π»
2. If π β πΊ, then ππ»πβ1 β π»
3. If π β πΊ, then ππ» = π»π
4. Every right coset is a left coset, i.e., if π β πΊ, β π β πΊ
with π»π = ππ».
Proof
1) β2) is obvious. Suppose 2) is true and show 3). We have (ππ»πβ1)a β π»π so ππ» β π»π. Also π(π-1 π»π) β ππ» so π»π βππ». Thus ππ» = π»π.
3) β4) is obvious. Suppose 4) is true and show 3). π»π =ππ» contains a, so ππ» = ππ» because a coset is an equivalence class. Thus ππ» = π»π.
Finally, suppose 3) is true and show 1). Multiply ππ» = π»π on the right by π-1.
Definition of normality for subgroups
Definition If π» satisfies any of the four conditions above, then π» issaid to be a normal subgroup of πΊ. (This concept goes back toEvariste Galois in 1831.)
Note For any group πΊ, πΊ and π are normal subgroups. If πΊ is anabelian group, then every subgroup of πΊ is normal.
Exercise Show that if π» is a subgroup of πΊ with index 2, then π» isnormal.
Quotient Groups
Theorem Suppose πΊ is a multiplicative group, π is a normal
subgroup, and πΊ/π is the collection of all cosets. Then
(ππ) . (ππ) = π(π . π) is a well defined multiplication (binary
operation) on πΊ/π, and with this multiplication, πΊ/π is a group. Its
identity is π and (ππ)β1=ππβ1 . Furthermore, if πΊ is finite,
π(πΊ/π) = π(πΊ)/π(π).
Proof Multiplication of elements in πΊ/π is multiplication of subsets
in πΊ . (ππ). (ππ) = π(ππ)π = π(ππ)π = π(π . π). Once
multiplication is well defined, the group axioms are immediate.
Example
Suppose G = π under +, n > 1, and N = nπ. Zn, the group of integersmod n is defined by πn = π/ππ. If a is an integer, the coset a +nZ is denoted by [a]. Note that [π] + [π] = [π + π], β[π] =[βπ], and [π] = [π + ππ] for any integer l. Any additive abeliangroup has a scalar multiplication over π, and in this case it is just[a]m = [am].
Note that [π] = [π] where r is the remainder of a divided by n, andthus the distinct elements of πn are [0], [1], . . . , [π β 1]. Also πn iscyclic because each of [1] and [β1] = [π β 1] is a generator.
We already know that if p is a prime, any non-zero element of Zp is agenerator, because Zp has p elements.
Criteria for the generator of cyclic group
Theorem If π > 1 and π is any integer, then [π] is a generator of πn iff (π, π) = 1.
Proof The element [π] is a generator iff the subgroup generated by [π] contains [1] iff β an integer π such that [π]π = [1] iff βintegers k and π such that ππ + ππ = 1.
Exercise Show that a positive integer is divisible by 3 iff the sum of its digits is divisible by 3. Note that [10] = [1] in Z3.
Homomorphisms
Homomorphisms are functions between groups that commute withthe group operations. It follows that they honor identities andinverses. In this section we list the basic properties.
Definition If πΊ and πΊ are multiplicative groups, a function π βΆ G β πΊis a homomorphism if, for all , b β πΊ, π a . b = π π . π π . Onthe left side, the group operation is in G, while on the right side it isin G.
The kernel of π is defined by ker π = πβ1 π = {a β πΊ βΆπ (π) = π}. In other words, the kernel is the set of solutions to theequation π (π₯) = π.
Examples
The constant map π βΆ G β πΊ defined by π (π) = π is ahomomorphism. If H is a subgroup of G, the inclusion i βΆ H β G isa homomorphism.
The function π βΆ π β π defined by π (π‘) = 2π‘ is ahomomorphism of additive groups, while the function defined byπ (π‘) = π‘ + 2 is not a homomorphism.
The function h βΆ π β π β 0 defined by β(π‘) = 2π‘ is ahomomorphism from an additive group to a multiplicative group.
If the homomorphism π βΆ πΊ β πΊ is a bijection, then the functionπβ1: πΊ β πΊ is a homomorphism. In this case, π is called anisomorphism, and we write πΊ β πΊ. In the case πΊ = πΊ π is alsocalled an automorphism.
Permutations
Definition Suppose π is a (non-void) set. A bijection π βΆ π β π is
called a permutation on π , and the collection of all these
permutations is denoted by π = π(π). In this setting, variables are
written on the left, i.e., π = (π₯)π. Therefore the composition
π π π means βπ followed by πβ. π(π) forms a multiplicative group
under composition.
Exercise Show that if there is a bijection between X and Y, there is
an isomorphism between π(π) and π(π). Thus if each of X and Y
has n elements, π (π) β π(π), and these groups are called the
symmetric groups on n elements. They are all denoted by the one
symbol ππ.
Permutations
Exercise Show that π(ππ) = π!. Let π = {1, 2, . . . , π}, πn = π(π), and
π» = {π β ππ βΆ (π)π = π}. Show π» is a subgroup of ππ which is
isomorphic to ππβ1. Let π be any permutation on π with
(π)π = 1. Find π-1π»π.
Cayleyβs Theorem
The next theorem shows that the symmetric groups are incredibly rich and complex.
Theorem Suppose πΊ is a multiplicative group with π elements and ππ is the group of all permutations on the set πΊ. Then πΊ is isomorphic to a subgroup of ππ.
Proof Let β βΆ πΊ β πn be the function which sends a to the bijection βa : πΊ β πΊ defined by (π)βa = π . π. The proof follows from the following observations.
For each given π, βa is a bijection from πΊ to πΊ.
β is a homomorphism, i.e., βππ = βa o βb.
β is injective and thus πΊ is isomorphic to image(β) β πn.
The Symmetric Groups
Now let π β₯ 2 and let πn be the group of all permutations on {1, 2, . . . , π}. The following definition shows that each element of πnmaybe represented by a matrix.
Definition Suppose 1 < π β€ π, {π1, π2, β¦ , ππ} is a collection of
distinct integers with 1 β€ ππ β€ π, and {π1, π2, β¦ , ππ} is the same
collection in some different order. Then the matrix
π1 π2β¦ πππ1 π2β¦ ππ
represents π β Sπ defined by (ππ)π = ππ for 1 β€ π β€
π, and (π)π = π for all other π . The composition of two
permutations is computed by applying the matrix on the left first and
the matrix on the right second.
Cycles
There is a special type of permutation called a cycle. For these we
have a special notation.
Definition π1 π2β¦ ππβ1πππ2 π3β¦ ππ π1
is called a π βcycle, and is denoted
by (π1, π2, β¦ , ππ).
A 2-cycle is called a transposition. The cycles (π1, π2, β¦ , ππ) and
π1, π2, β¦ , ππ are disjoint provided ππ β ππ for all 1 β€ π β€ π and 1 β€ j
β€ l.
Basic properties of permutations
Listed here are basic properties of permutations. They are all easy except 4) which takes a little work. Properties 9) and 10) are listed solely for reference.
Theorem
1. Disjoint cycles commute. (This is obvious.)
2. Every non-identity permutation can be written uniquely (except for order) as the product of disjoint cycles. (This is easy.)
3. Every permutation can be written (non-uniquely) as the product of transpositions. (Proof: πΌ = (1, 2)(1, 2) and (π1 , β¦ , ππ) = (π1, π2 )(π1, π3) β¦ (π1ππ). )
4. The parity of the number of these transpositions is unique. π is the product of π transpositions and also of q transpositions, then π is even iff π is even. In this case, π is said to be an evenpermutation. In the other case, π is an odd permutation.
Basic properties of permutations
5. A π βcycle is even (odd) iff π is odd (even). For example (1, 2, 3) = (1, 2)(1, 3) is an even permutation.
6. Suppose π, π β πn. If one of f and g is even and the other is odd, then π o π is odd. If π and g are both even or both odd, then π π π is even. (Obvious.)
7. The map β βΆ πn β π2 defined by β(even) = [0] and β(odd) = [1]is a homomorphism from a multiplicative group to an additive group. Its kernel (the subgroup of even permutations) is denoted by π΄n and is called the alternating group. Thus π΄n is a normal subgroup of index 2, and ππ/π΄π β π2.
Basic properties of permutations
8. If π, π, π and d are distinct integers in {1, 2,..., π} , then(π, π)(π, π) = (π, π, π) and (π, π)(π, π) = (π, π, π)(π, π, π) .Since πΌ = (1, 2, 3)3, it follows that for π β₯ 3 , every evenpermutation is the product of 3-cycles.
The following parts are not included in this course. They arepresented here merely for reference.
9. For any π β₯ 4, π΄ n is simple, i.e., has no proper normalsubgroups.
10. ππ can be generated by two elements. In fact,{(1, 2), (1, 2, . . . , π)} generates ππ . (Of course there aresubgroups of ππ which cannot be generated by two elements).
Exercises
1. Write1 2 3 4 5 6 76 5 4 3 1 7 2
as the product of disjoint cycles.
2. Write (1,5,6,7)(2,3,4)(3,7,1) as the product of disjoint cycles.
3. Write (3,7,1)(1,5,6,7)(2,3,4) as the product of disjoint cycles.Which of these permutations are odd and which are even?
4. Suppose (π1, π2, β¦ , ππ) and π1, π2, β¦ , ππ disjoint cycles. What isthe order of their product?
5. Suppose π β ππ . Show that Οβ1 1, 2, 3 π =((1) π, (2) π, (3) π). This shows that conjugation by Ο is just atype of re-labeling. Also let π = (4, 5, 6) and findπβ1 (1, 2, 3, 4, 5)π.
6. Show that π» = { π β π6: (6)π = 6} is a subgroup of π6 andfind its right cosets and its left cosets.