Example: Ifthenis an automorphism of the group of complex numbers under addition. We test the requirements one by one.
1. With
2. Ifthenandsois one to one.
3.is onto since ifthenand
4.
The mappingsandare similarly automorphisms. All these automorphisms are length preserving.
A very important automorphism is the inner automorphism,whereis some element ofThis is called the automorphism ofinduced by
The inner automorphism ofinduced by(rotation by) is shown below.
The set of inner automorphisms is a group, as is the set of automorphisms.
]]>Formally, let G be our group, with operation the group operation. Let C be the Cayley table for the group, withdenoting the element at rowand columnwhich is the result of applying elementwith
Example: The Cayley table for the groupis shown below.
The group has elements (0,1,2,3,4). Each element appears once along the top row and down the left hand side, and once in the body of the table. The identity element 0 is symmetrically distrubuted, and this is so whether the group is Abelian so thator non – AbelianThe above group is Abelian so the whole table is symmetric. The Cayley table forthe group of symmetries of a square is shown below.
All Cayley tables for isomorphic groups are isomorphic (that is, the same, invariant of the labeling and ordering of group elements).
It has all of the features of the first table with the exception of being symmetric.is non – Abelian so the Cayley table is not symmetric.
]]>Ifis a group of orderthenis isomorphic to a subgroup of the permutation group
Example: Letbe the Klein groupof order 4,with every element self inverse, so of order 2.
Example: We can definethenis an isomorphism fromonto
Proof of Cayley's Theorem
Let theelements ofbeFor an elementwe find the permutation of the elements offormed by multiplying each element on the left byEachmust be equal tofor someif and only if
We must show
is a uniquely defined element offor each
is one to one.
has the morphism property,
By constructionmaps the setto itself.is one to one since ifand thenandand 2 is proved. Hence is a one to one map of the setto itself , so is a permutation of this set andso 1 is satisfied.
To show 3, we show thatandhave the same effect on each of the elements in
Assumeandso that
By the definition ofif and only ifandif and only ifby associativity inbutif and only iffor eachhenceandhave the same effect on every element ofsoand Cayley's Theorem is proved.
]]>For groups up to order 8 these are:
Order | Groups |
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | |
9 | |
10 | |
11 | |
12 | |
13 | |
14 | |
15 |
Various methods exist for finding these groups – using Sylow's theorems, analysis of the conjugacy class equation, Lagrange's equation.
]]>Conjugacy is an equivalence relation since
so
so
so
splits into equivalence classesEvery element of the group belongs to precisely one conjugacy class, and the classesandare equal if and only ifandare conjugate, and disjoint otherwise. The equivalence class that contains the elementisand is called the conjugacy class ofThe class number ofis the number of distinct (nonequivalent) conjugacy classes.
Elements of each conjugacy class have a similar structure. If the group elements act on a geometric object, elements of each conjugacy class have similar geometric effects. For example, all the rotations may form one class, all the reflections another, and the identity will be in a class by itself.
The symmetric groupconsisting of all 6 permutations of three labels, has three conjugacy classes:
The identity (1)(2)(3).
interchanging two two labels (12)(3), (13)(2),(1)(23).
a cyclic permutation of all three labels (123), (132).
Ifis abelian, thenfor all sofor allso conjugacy is not very useful in the abelian case. A subset of the group may be abelian so the conjugacy classes gives us an idea of the extent of non – abelianness.
Ifbelong to the same conjugacy class - they are conjugate - then they have the same order and every statement aboutcan be translated into a statement aboutbecause the mapis an automorphism of
An elementlies in the centerofif and only if its conjugacy class has only one element, a itself. More generally, ifdenotes the centralizer ofi.e., the subgroup consisting of all elementssuch thatthen the indexis equal to the number of elements in the conjugacy class of(by the orbit-stabilizer theorem).
Ifare conjugate, then so are powers of them,since
]]>Ifor equivalentlyfor allthenis a normal subgroup ofThe left and right cosets coincide and the set of cosets forms a group with the group operation defined byForunder the operation addition modulo 3, the cosets ofareandIf G is abelian then all subgroups are normal.
if and only ifis an element of H since as H is a subgroup, it must be closed and must contain the identity.
Any two left cosets ofinare either identical or disjoint — i.e., the left cosets form a partition ofsuch that every element ofbelongs to one and only one left coset. In particular the identity is in precisely one coset, and that coset isitself since this is also the only coset that is a subgroup.
The left cosets ofinare the equivalence classes under the equivalence relation ongiven byif and only ifand similarly for right cosets.
All left cosets and all right cosets have the same order (number of elements, or cardinality in the case of an infinite H), equal to the order of H (because H is itself a coset). Furthermore, the number of left cosets is equal to the number of right cosets and is known as the index of H in G, written asLagrange's theorem allows us to compute the index in the case whereandare finite, as per the formula:
For example, ifis a group, thenand G is cyclic. In fact,is isomorphic towith additionFor example, corresponds toWe can use the isomorphismdefined by
For every positive integerthere is exactly one cyclic group (up to isomorphism) whose order isand there is exactly one infinite cyclic group (the integers under addition). Hence, the cyclic groups are the simplest groups and they are completely classified.
Since the cyclic groups are abelian, they are often written additively and denotedor or C-n where n is the order, equal to the number of elements.
inwhereas 3 + 4 = 2 in
Cyclic groups and all their subgroups are abelian. Every element is of the formthen so every element commutes with every other.
Ifthenfor allThis is because
Ifis a cyclic group of orderthen every subgroup ofis cyclic. The order of any subgroup ofis a divisor ofand for each positive divisorofthe grouphas exactly one subgroup of order
Ifis finite, then there are exactlyelements that generate the group on their own, whereis the number of numbers inthat are coprime toMore generally, ifdividesthen the number of elements inwhich have orderisThe order of the residue class of m is
Ifis prime, then the only group (up to isomorphism) with p elements is the cyclic groupor
The direct product of two cyclic groupsandis cyclic if and only ifand are coprime. Thusis the direct product ofandbut not the direct product ofand
A primary cyclic group is a group of the formwhereis a prime number. The fundamental theorem of abelian groups states that every finitely generated abelian group is the direct product of finitely many finite primary cyclic and infinite cyclic groups.
The elements ofcoprime toform a group under multiplication modulowith elements,Whenwe get
is cyclic if and only ifforandin which case every generator ofis called a primitive root moduloThus,is cyclic for but not forwhere it is instead isomorphic to the Klein four-group.
The group is cyclic withelements for every primeand is also written because it consists of the non-zero elements. More generally, every finite subgroup of the multiplicative group of any field is cyclic.
]]>Ifis a collection of groups, the external direct productis the set of all n – tuples with the ith component is an element ofWe can writeand define the product of elements ofasThe productis done using the group operation of
The direct product of any number of groups is itself a group. If the order ofis m-i then the order ofis
An obvious example of a direct product isEach component ofis a real number.is a group with the group operation being addition, sois a group with the group operation being componentwise addition.
is a group with components first component one offrom and the second component one offrom
If all theare abelian, then so is
If theare cyclic of orderrespectively, and none of thehave any common factors, thenis cyclic of order
]]>The most common way to formalize this is by defining a field as a set together with two operations, usually called addition and multiplication, and denoted byandrespectively, such that the following axioms hold;
Note that all but the last axiom are exactly the axioms for a commutative group, while the last axiom is a compatibility condition between the two operations.
Examples:are all fields as isfor n prime. The tables below are for a finite field with four elements.
0 | 1 | A | B | 0 | 1 | A | B | |||
0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | A | B | |
1 | 0 | 1 | A | B | 1 | 1 | 0 | B | A | |
A | 0 | A | B | 1 | A | A | B | 0 | 1 | |
B | 0 | B | A | 1 | B | B | A | 1 | 0 |
The coefficientsare rational, and we have only used multiplication, division, addition, subtraction and square root.
We can find more complicated examples, suppose p(x)=x^4 +4x^2+2. We can write this asso the solutions will satisfy The roots are
When we can find the solutions for a polynomial with rational coefficients using only rational numbers and the operations of addition, subtraction, division, multiplication and finding nth roots, we say thatis soluble by radicals.
Using Galois theory, we can prove that if the degree of(the highest power ofin) is less than 5 then the polynomial is soluble by radicals, but there are polynomials of degree 5 and higher not soluble by radicals. In other words, polynomials of degree 5 whose solutions cannot be written down using nth roots and the arithmetical operations, no matter how complicated.
We can construct a group to act of the set of roots of a polynomial – called a group action. Such a group will be an automorphism of the roots. For example the group acting on the roots of the polynomialare
and
For a polynomial of degreethe group will be a subgroup ofThe group generated will have subgroups which may or may not be normal inIf the subgroup is normal inthe the polynomial is soluble by radicals else it is not. Forandall the subgoups are normal butandforhas subgroups which are not normal, so polynomials of degree 5 or greater are not soluble by radicals in general although some may be.
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