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1 Contents 1 Groups Examples of Groups Things that are not groups Properties of Groups Rings and Fields Examples Some Finite Fields Introduction to Vector Spaces Denition of a Vector Space Examples of Vector Spaces Linear Independence and Bases Orthogonality and Null Spaces 15

2 2 1 GROUPS 1 Groups In this section we introduce the concept of an abstract algebra. We are all familiar with algebra from high school, maybe even primary school. The idea of abstract algebra is that we can dene algebraic systems, based on particular sets of axioms and then build up theorems starting only from the axioms. In this way, the theorems remain true in any system that satises the axioms. One way of thinking of this is \object oriented mathematics". For most (engineering) students, abstract algebra is the rst contact with a rigorous axiom-theorem-proof type of mathematical system. While the material and concepts are simple, the methodology can feel constricting, even pedantic, as there is no appeal to outside help or intuition. Every theorem can be proved using only existing theorems, and the given axioms. In abstract algebra, we are interested in sets, and operations on elements in sets. We shall not make the mistake of trying to dene a set. We will however have some basic assumptions. The rst algebraic system that we shall consider is the group, which consists of a set and a binary operation on elements from the set, obeying certain axioms. Denition 1.1 (Group) A group <G;>is a set G and a binary operation such that the following axioms are satised. G1. (Closure) For any a; b 2 G, the result of a b is also in G. G2. (Associativity) For a; b; c 2 G, (a b) c = a (b c): G3. (Identity) There exists an element e 2 G, such that e x = x e = x for all x 2 G. This element e is called the identity element for <G;>. G4. (Inverses) For every element a 2 G, there exists another element b 2 G which has the property a b = b a = e: The element b is called the inverse of a in <G;>.

3 1.1 Examples of Groups 3 Note that because of our denition of binary operation, axiom G1 is actually redundant. We include it just to re-inforce the fact that a group is closed. It is also important to note that commutativity is not required for a group. This means that in a group, we do not have tohaveab=ba. While that might seem strange, there is an example of such a group that you are already familiar with. If the binary operation of a group is commutative, the group has a special name Denition 1.2 (Abelian Group) An abelian group is a group in which the binary operation is commutative. That is a b = b a. Abelian groups are named after Niels Abel. A group is not the simplest algebraic structure, but it is the simplest one that we shall need. Just in case you come across these terms in your reading, we provide for reference the following denitions: Semigroup. A semigroup is a set G and a binary operation such that G1 and G2 hold. Monoid A monoid is a semigroup with an identity element, i.e. G1, G2 and G3 hold. 1.1 Examples of Groups Let us now look at some examples of groups. We shall use the following common notations: Zis the set of all integer numbers. Q is the set of all rational numbers a=b where a; b 2 Z. R is the set of all real numbers. C is the set of all complex numbers. We use a superscript + to indicate that one of these sets is restricted to positive values only, e.g. Z + is the set of all positiveintegers. Note that zero is not a positive number.

4 4 1 GROUPS Example 1.1 (Integers, Addition) Z, under addition is a group. the identity is 0, and the inverse of any element a 2 Zis,a. The group properties follow from the properties of the integers. This is an abelian group. Example 1.2 (Addition) R, Q and C are all abelian groups under addition. Example 1.3 (Multiplication) R +, Q + and C + are all abelian groups under addition. Example 1.4 (Matrices, Addition) The set of all n n matrices under matrix addition is an abelian group. What is the identity element? Example 1.5 (Matrices, Multiplication) The set of all nn non-singular real matrices is a group under matrix multiplication. This is a non-commutative group. What is the identity element? Example 1.6 (Modulo-m Addition) Let G = f0; 1; 2;:::; m,1g, and for every a; b in G, dene a b to be the modulo-m addition of a and b. Then <G;>is a group. We are particularly interested in this group. You may be already familiar with this group for m =2. In this case, it is just 0 and 1, under normal \binary addition", or exclusive or operation. Example 1.7 (Modulo-m Multiplication) Let G = f1; 2;:::; p,1g, and for every a; b in G, dene a b to be the modulo-p multiplication of a and b. Then provided p is a prime number, <G;>is a group. If p is not prime, G1 is not satised. 1.2 Things that are not groups The following examples are of sets and operations that while sounding plausible, are not groups. Example 1.8 (Integers, Multiplication) Z, under multiplication is not a group. Why not? Example 1.9 (Positive Integers) Z + under addition is not a group. Why not? This is an example of a semigroup.

5 1.3 Properties of Groups 5 Example 1.10 (Multiplication) R, Q and C under multiplication are not groups. Why not? Example 1.11 (Matrices, Multiplication) The set of all n n matrices under matrix multiplication is not a group. Why not? 1.3 Properties of Groups Here are some examples that we looked at in class, just to give aavour of working with groups. The whole point is that once the theorems are developed for groups, they hold for any algebraic system that satises the group axioms. Theorem 1.1 (Cancellation) In a group < G; >, the left and right cancellation laws hold, that is a b = a c implies b = c and b a = c a implies b = c. Proof: We will prove the left cancellation law. Suppose a b = a c. Then by G4 there exists an element d such that d a = e. Hence Now using G2, we know But d is the inverse of a, and From the denition of e, wehaveb=c. d (a b) =d(ac): (d a) b =(da)c: e b = e c: Theorem 1.2 (Unique Identity) The identity element e 2 G is unique. Proof: Suppose that e is not unique. Then there exists another element, say g with the property that for all a 2 G, g a = a. Then e = g e = g, using the properties of e and g. Theorem 1.3 (Unique Inverses) The inverse for each element a 2 G is unique.

6 6 2 RINGS AND FIELDS Proof: Suppose there are two elements, b and c with the property b a = e and c a = e. Then b a = c a, and by the right cancellation theorem b = c. Theorem 1.4 (Linear Equations) Let a; b be elements of a group < G; >. Then the linear equation a x = b has a unique solution for x in G. Proof: First we show that if c is the inverse of a (which exists because of G4), x = c b is a solution. Now, substituting this in for x on the left hand side, a (c b) =(ac)b G2 = e b Def. of c = b G3 hence x = c b is a solution. Now suppose that there is another solution, say y. Then a x = a y, and by the cancellation law x = y. Thus the solution is unique. 2 Rings and Fields Although groups are very interesting and useful in their own right, we are probably more comfortable in algebraic systems that have not one, but two binary operations (which wemaybe tempted to call addition and multiplication). We shall now introduce algebraic structures with two operations. Such structures are useful in a fundamental area of mathematics: solving polynomial equations. Denition 2.1 (Ring) A ring <R;;>is a set R and two binary operations, and, dened on R, which we call addition and multiplication, such that the following axioms are satised. R1. <R;>is an abelian group. R2. For any a; b; c 2 R, wehave(ab)c=a(bc), i.e. multiplication is associative. R3. For all a; b; c 2 R, wehave a(bc)=(ab)(ac) (bc)a=(ba)(ca); I.e. multiplication distributes over addition, either from the right or the left.

7 7 Note that R2 implies that <R;>must be at least a semigroup. It does not however have to have an identity element, inverses or be commutative. If it does, the we have our next structure. Theorem 2.1 (Properties of Rings) If R is a ring with additive identity 0, then for any a; b 2 R, wehave 1. 0 a = a 0=0, 2. a (,b) =(,a)b=,(ab), 3. (,a) (,b) =ab. Proof: 1. By R1 and R2 we have 0a+0a=(00)a =0a Hence by the cancellation law for the additive group < R; >,wehave 0 a= 0. Proof for right multiplication by zero is similar. 2. We need to prove that a (,b) =,(ab), or alternatively that a (,b)+ (ab) = 0, since by denition,,(a b) is the element ofrwhich when added to (a b) gives 0. But a (,b)+ab=a(,bb) Proof for the remaining part is similar. =a0 =0: 3. From the preceding property, note that (,a) (,b) =,(a(,b)). A second application yields,(a (,b)) =,(,(a b)), which by denition is the element of R which when added to,(a b) gives 0. This element is of course a b.

8 8 2 RINGS AND FIELDS Denition 2.2 (Field) Let <F;;>be a ring, which in addition to R1-R3, obeys the following axioms F1-F3. Then <F;;>is called a eld. F1. is commutative. F2. There is an (unique) element F, denoted by 1 with the property 1 a = a = a 1; for all a 2 F. This element is the multiplicative identity, or unity. F3. Every element in F, except for the additive identity, 0 has a multiplicative inverse. For every a 2 F there exists a b 2 F such that a b =1=ba: Note: The multiplicativeinverse for an element ais sometimes denote a,1,or 1=a. Denition 2.3 (Subeld, Extension Field) Let <E;;>be a eld, and let F E be such that <F;;>is also a eld. Then F is a subeld of E, and E is an extension eld of F. Fields are quite powerful objects, because everything that we know about solving polynomial equations with real or complex coecients translates directly into equations with coecients from arbitrary elds. In particular, we can solve systems of linear equations using matrix methods, where now we have matrices with elements from elds other than the reals or complex numbers. Because of this, when we are not being especially pedantic, we replace the notation with + and with : or simply specify multiplication by juxtaposition. 2.1 Examples You may thing that the idea of a ring is pretty esoteric, and that a eld sounds more like real-life. The following example would challenge you about this. Example 2.1 (Integers) Zis a ring (with multiplicative unity). It is not however a eld, since only 1 and,1 have multiplicative inverses.

9 2.1 Examples 9 Example 2.2 (Common Fields) It is an exercise to check that Q;R and C are all elds. For us, the most important example of a eld is a em nite eld, that is one in which F is a nite set. Example 2.3 (Finite Fields) Let F =0;1;2;:::;p,1, where p is a prime number. Let and be modulo-p addition and modulo-p multiplication respectively. Then <F;;>is a eld. Such elds are called Galois Fields, in honour of their discoverer, Evariste Galois. We shall denote these elds by GF(p). Theorem 2.2 The set F = 0;1;2;:::;p,1, with binary operations and dened by modulo-p addition and modulo-p multiplication is a eld if and only if p is prime. Proof: We begin the proof by showing that a eld can have nodivisors of zero, by which we mean that the product of two elements can be zero if and only if one of the elements is zero. Assume that < F;; >is a eld, and let a; b 2 F with a 6= 0besuch that a b = 0. Then (by F3), we have This in turn implies (because of R2) a,1 (ab)=a,1 0: (a,1 a) b =0; and hence b = 0. Hence we have proved that a eld can have no divisors of zero. Now, if p is not prime, there are non-zero elements m; n 2 F such that m n = p, which is eqivalent to zero modulo p. This implies that divisors of zero would exist if p were not prime. It is one of the nicer results of abstract algebra that there is essentially only one nite eld for each prime order. Without going into details, all elds of order p have basically the same structure.

10 10 2 RINGS AND FIELDS 2.2 Some Finite Fields Here are some examples of nite elds. Remember that the set f0; 1; 2;:::;p,1g under modulo-p addition and multiplication only form a eld if p is prime. First, let us look at the \addition" and \multiplication" tables for GF(2) and GF(3). Example 2.4 (GF(2)) Example 2.5 (GF(3)) Now lets look at GF(4). The number 4 is not prime. This does not mean that GF(4) does not exist, only that it is not dened by modulo-4 addition and multiplication, as we shall now see. You can verify that the tables below indeed give a eld. Example 2.6 (GF(4)) It is very important not to get confused by the labels and in the above tables. They are not modulo-4 addition and multiplication, just some operations that give a eld. Perhaps it would be better to re-label everything so that it doesn't look like numbers any more: Example 2.7 (GF(4)) Let G = 1;!g. <G; ;>is a eld dened as follows:

11 1! ! 1! 1 Structurally, this is exactly the same as before, we have just re-labelled everything. You may wonder how to nd such elds as GF(4). Its not magic, there is a way to construct elds GF(p m ), where p is prime. We shall cover this in a later lecture. 3 Introduction to Vector Spaces Most of us are familiar with the concept of a vector space from elementary linear algebra. This includes the concepts of vectors (n-tuples of real or complex numbers), scalars (real or complex numbers) and binary operations involving these objects, such as adding two vectors, or multiplying a vector by a scalar. These concepts however are not restricted to n-tuples of numbers. A vector space may be dened in a more abstract way, retaining the main ideas such as span, bases and dimension. 3.1 Denition of a Vector Space We now give a formal denition of a vector space, which is basically a eld, a group and a scalar multiplication operation, obeying certain rules. For clarity, we have used dierent symbols for eld operations, group operations and scalar multiplication. The operation in question however can always be determined by the operands, and usually no distinction is made. Denition 3.1 (Vector Space) Let <F;;>be a eld. A vector space over F consists of an abelian group < V;+> and a scalar multiplication operation, f v dened for each f 2 F and v 2 V, such that for all a; b 2 F and v;w 2 V, the following conditions hold. V1. (Closure) a v 2 V. The result multiplying a scalar and a vector is always a vector.

12 12 3 INTRODUCTION TO VECTOR SPACES V2. (Associativity) a(bv) =(ab)v. The order of operation in multiplicative expressions doesn't matter. V3a. (Distributivity) (a b) v = a v + b v. Vectors distribute over scalar addition. V3b. (Distributivity) a (v + w) = a v + a w. Scalars distribute over vector addition. V4. (Unit scaling) 1 v = v. The scalar 1 acts as a multiplicativevector identity. Denition 3.2 (Subspace) Let < V;+> be a vector space over F. Then U V is a subspace of V if and only if <U;+>is a vector space over F (with the same scalar multiplication). A subset S V ofavector space over the eld F is a subspace if and only if for every v 1 ;v 2 2 S, and every a; b 2 F, the linear combination a v 1 + b v 2 2 S. Abstract vector spaces obey some simple relationships that we are probably already familiar with. Let c and v be a scalar and vector respectively. The vector 0 denotes the additive identity in <V;+ >. Then 0 v =0 c0=0 (,c)v=,(cv) 3.2 Examples of Vector Spaces As an exercise, you can check that the following examples obey conditions V1-V4.

13 3.2 Examples of Vector Spaces 13 Example 3.1 (Real n-vectors) Let V =< R n ; + > be the set of all real n- tuples under component-wise addition. For v =(v 1 ;v 2 ;:::;v n ) 2 V, dene scalar multiplication by r 2 R as r v =(rv 1 ;rv 2 ;:::;rv n ); where on the right we have the usual real multiplication. Then V is a vector space over R. This is the vector space that you are probably already familiar with from linear algebra. Example 3.2 (n-tuples over F ) For any eld <F;;>, let V =< F n ;+> be the set of all n-tuples with elements from F, where vector addition is componentwise addition in F. Then V is a vector space over F, with scalar multiplication dened for v =(v 1 ;v 2 ;:::;v n )2 V, and f 2 F as f v =(fv 1 ;f v 2 ;:::;f v n ): This is probably the most important example for us in error control coding. We will dene a linear block code to be simply a subspace of GF(2) n. The next gure shows an example of a 3 dimensional vector space over GF(2). The vector space consists of the vertices of the cube. Also shown is a subspace, consisting of the vectors f000; 001; 010; 011g Figure 1: Vector space over GF(2).

14 14 3 INTRODUCTION TO VECTOR SPACES The next few examples may sound a bit more sophisticated, but the principle involved is the same. Example 3.3 (Polynomials) For any eld F, let F [x] be the set of all polynomials in x, with coecients in F. V =< F[x];+>, the group of polynomials under normal addition of polynomials is a vector space over F, where the scalar multiplication is just multiplication in F of the coecients by the scalar. Example 3.4 (Real Functions) The set of all real functions on some support (with addition dened in the usual way) form a vector space over R, where scalar multiplication of a function f(x) by a scalar c is dened by the real multiplication cf(x). Just for reference, we give the following example. We will cover extension elds when we look at cyclic codes. Example 3.5 (Extension Fields) Let E be an extension eld of F. Then E is avector space over F, with vector addition dened by addition in E, and scalar multiplication dened by multiplication in E. 3.3 Linear Independence and Bases Denition 3.3 (Linear Combination) For an integer n, scalars c 1 ;c 2 ;:::;c n and vectors v 1 ;v 2 ;:::;v n, the vector nx i=1 is a linear combination of the v i. c i v i = c 1 v 1 + c 2 v 2 + :::c n v n Denition 3.4 (Linear Independent) The vectors v 1 ;v 2 ;:::;v n are said to be linearly independent if and only if nx i=1 c i v i =0 =) c i =0; for all i =1;2;:::;n. Otherwise, they are said to be linearly dependent.

15 15 Denition 3.5 (Span) For an integer n, let v 1 ;v 2 ;:::;v n be vectors in some vector space V over the scalar eld F. Then the set of all possible linear combinations of the v i, namely S = ( n X i=1 c i v i : c i 2 F is a subspace of V and is called the span of the vectors v 1 ;v 2 ;:::;v n. This is denoted by S = span(v 1 ;v 2 ;:::;v n ). Denition 3.6 (Finite Dimension) Avector space V is said to be nite dimensional if there is a nite subset of V that spans V. Denition 3.7 (Basis) If V isavector space over F, and V = span(v 1 ;v 2 ;:::;v n ) for some linear independent set of vectors B = fv i gv, then B is said to form a basis for V. Theorem 3.1 (Dimension) Let V be a nite dimensional vector space. Then the number of elements in a basis for V is the same, regardless of the choice of these vectors. This number is called the dimension of V, denoted dim(v ). Theorem 3.2 (Unique Representation) Let B = fv 0 ;v 1 ; ;v n,1gbe a basis for the n-dimensional vector space V. Then every element of V has a unique representation as a linear combination of the basis vectors. ) ; 4 Orthogonality and Null Spaces In error control coding, we are mainly interested in the vector space F n of n-tuples over some nite eld F. In addition to vector addition, we can dene an inner product (sometimes called a dot product) for such n-tuples. Denition 4.1 (Inner Product) Let v;w 2 F n. The inner product v w : F n F n 7! F is dened v w = v 1 w 1 v 2 w 2 :::v n w n :

16 16 4 ORTHOGONALITY AND NULL SPACES You should recognise this inner product from linear algebra, where you would have encountered it for real vectors. If v and w are written as column vectors, we have vw=v T w. Denition 4.2 (Orthogonal) Two vectors v;w 2 F n are orthogonal if and only if v w =0: Note that in nite eld arithmetic, a non-zero vector can be orthogonal to itself, something that is impossible over the reals. Can you think of a simple example of this? Denition 4.3 (Null Space) If S is a subspace of F n, the set of vectors in F n that are orthogonal to every vector in S form a subspace, called the null space of S. The null space of S in F n is denoted by S?. S? = fv 2 F n : v w =0;8w2Sg This is also known as the orthogonal subspace of S in F n. These orthogonal subspaces have some very nice properties. Theorem 4.1 (Total Dimension) Let S F n be a subspace of V, with dim(s) = k. Then dim(s? )=n,k. Theorem 4.2 (S? )? = S. Example 4.1 Let S = f0000; 0101; 0001; 0100g be a subspace of GF(2) 4. Then S? = f0000; 1010; 1000; 0010g.

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