# Lect3-02web

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MATH 415 Modern Algebra I Lecture 14: Follow-up on Exam 1. Advanced algebraic structures.
Follow-up on Exam 1 Problem 1a. Consider a binary operation on R given by x y = 3 radicalbig x 3 + y 3 for any x , y R . Is ( R , ) a semigroup? Is it a group? First we check that the operation is well defined: x , y R = x y R . Then we observe that ( x y ) 3 = x 3 + y 3 for all x , y R . Consider a function f : R R given by f ( x ) = x 3 . We obtain that f ( x y ) = f ( x ) + f ( y ) for all x , y R . This means that f is a homomorphism of the binary structure ( R , ) to the binary structure ( R , +). It is easy to see that f is bijective. It follows that ( R , ) is a group and that it is isomorphic to the group ( R , +).
Problem 1b. Consider a binary operation on R given by x y = y for all x , y R . Is ( R , ) a semigroup? Is it a group? Clearly, x y R for all x , y R so that the operation is well defined. Further, for any x , y , z R we have ( x y ) z = y z = z and x ( y z ) = x z = z . Hence ( x y ) z = x ( y z ), the operation is associative. By definition of the operation, every element is a left identity. It follows that there is no right identity (and hence this is not a group). Indeed, if e is any left identity and e r is any right identity, then e r = e e r = e .
Problem 3b. We have the following information on a group G and its subgroups H 1 and H 2 : the order of G is either 15 or 20 or 25, H 1 and H 2 are proper subgroups of different orders and neither of them contains the other, the intersection H 1 H 2 is a nontrivial subgroup of G . Find the order of H 1 H 2 .
Let n denote the order of the group G , let m 1 and m 2 denote orders of the subgroups H 1 and H 2 , and let k denote the order of the intersection H 1 H 2 . We are given that n =15 or 20 or 25, that m 1 < n and m 2 < n , that m 1 negationslash = m 2 , and that k > 1. By Lagrange's Theorem, the order of a subgroup divides the order of the group. In particular, m 1 and m 2 are proper divisors of n . Since neither of the subgroups H 1 and H 2 contains the other, their intersection H 1 H 2 is a proper subgroup of H 1 and of H 2 . By Lagrange's Theorem, k is a proper divisor of m 1 and of m 2 . Since k > 1, it follows that neither m 1 nor m 2 is a prime number or 1. Now we obtain that n cannot be 15 or 25 as any proper divisor of these numbers is either prime or 1. Therefore n =20. The number 20 has only two proper divisors which are composite numbers: 4 and 10. Hence one of them is m 1 and the other is m 2 . Finally, k is a common divisor of 4 and 10. Since k > 1, we obtain that k =2. Remark. The group G in this problem can be, for example, Z 2 × Z 2 × Z 5 . Then H 1 = Z 2 × Z 2 × { 0 } , H 2 = { 0 } × Z 2 × Z 5 and H 1 H 2 = { 0 } × Z 2 × { 0 } .
Problem. Find two non-abelian groups of order 24 that are not isomorphic to each other. It is known that groups of order 24 form 15 isomorphism classes. Three of them are abelian groups, represented by Z 3 × Z 8 , Z 3 × Z 4 × Z 2 , and Z 3 × Z 2 × Z 2 × Z 2 . The other 12 classes are non-abelian groups. Representatives for some of them are: S 4 , A 4 × Z 2 , S 3 × Z 4 , S 3 × Z 2 × Z 2 , D 12 , D 4 × Z 3 , and SL (2 , Z 3 ).
Rings Definition. A ring is a set R , together with two binary operations usually called addition and multiplication and denoted accordingly, such that R is an abelian group under addition, R is a semigroup under multiplication, multiplication distributes over addition. The complete list of axioms is as follows: (A0) for all x , y R , x + y is an element of R ; (A1) ( x + y )+ z = x +( y + z ) for all x , y , z R ; (A2) there exists an element, denoted 0, in R such that x +0=0+ x = x for all x R ; (A3) for every x R there exists an element, denoted x , in R such that x +( x )=( x )+ x =0; (A4) x + y = y + x for all x , y R ; (M0) for all x , y R , xy is an element of R ; (M1) ( xy ) z = x ( yz ) for all x , y , z R ; (D) x ( y + z )= xy + xz and ( y + z ) x = yx + zx for all x , y , z R .
Fields Definition. A field is a set F , together with two binary operations called addition and multiplication and denoted accordingly, such that F is an abelian group under addition, F \ { 0 } is an abelian group under multiplication, multiplication distributes over addition. In other words, the field is a commutative ring with unity (1 negationslash = 0) such that any nonzero element has a multiplicative inverse. Examples. Real numbers R . Rational numbers Q . Complex numbers C . Z p : congruence classes modulo p , where p is prime. R ( X ): rational functions in variable X with real coefficients.
Vector spaces over a field Definition. Given a field F , a vector space V over F is an additive abelian group endowed with a mixed operation φ : F × V V called scalar multiplication or scaling . Elements of V and F are referred to respectively as vectors and scalars . The scalar multiple φ ( λ, v ) is denoted λ v . The scalar multiplication is to satisfy the following axioms: (V0) for all v V and λ F , λ v is an element of V ; (V1) λ ( v + w ) = λ v + λ w for all v , w V and λ F ; (V2) ( λ + μ ) v = λ v + μ v for all v V and λ, μ F ; (V3) λ ( μ v ) = ( λμ ) v for all v V and λ, μ F ; (V4) 1 v = v for all v V . (Almost) all linear algebra developed for vector spaces over R can be generalized to vector spaces over an arbitrary field F . This includes: linear independence, span, basis, dimension, determinants, matrices, eigenvalues and eigenvectors.
Examples of vector spaces over a field F: The space F n of n -dimensional coordinate vectors ( x 1 , x 2 , . . . , x n ) with coordinates in F . The space M n , m ( F ) of n × m matrices with entries in F . The space F [ X ] of polynomials p ( x ) = a 0 + a 1 X + · · · + a n X n with coefficients in F . Any field F that is an extension of F (i.e., F F and the operations on F are restrictions of the corresponding operations on F ). In particular, C is a vector space over R and over Q , R is a vector space over Q .
Linear independence over Q Since the set R of real numbers and the set Q of rational numbers are fields, we can regard R as a vector space over Q . Real numbers r 1 , r 2 , . . . , r n are said to be linearly independent over Q if they are linearly independent as vectors in that vector space. Example. 1 and 2 are linearly independent over Q . Assume a · 1 + b 2 = 0 for some a , b Q . We have to show that a = b = 0. Indeed, b = 0 as otherwise 2 = a / b , a rational number. Then a = 0 as well. In general, two nonzero real numbers r 1 and r 2 are linearly independent over Q if r 1 / r 2 is irrational.
Linear independence over Q Example. 1, 2, and 3 are linearly independent over Q . Assume a + b 2 + c 3 = 0 for some a , b , c Q . We have to show that a = b = c = 0. a + b 2 + c 3 = 0 = a + b 2 = c 3 = ( a + b 2) 2 = ( c 3) 2 = ( a 2 + 2 b 2 3 c 2 ) + 2 ab 2 = 0. Since 1 and 2 are linearly independent over Q , we obtain a 2 + 2 b 2 3 c 2 = 2 ab = 0. In particular, a = 0 or b = 0. Then a + c 3 = 0 or b 2 + c 3 = 0. However 1 and 3 are linearly independent over Q as well as 2 and 3. Thus a = b = c = 0.
Finite fields Theorem 1 Any finite field F has nonzero characteristic. Proof: Consider a sequence 1 , 1+1 , 1+1+1 , . . . Since F is finite, there are repetitions in this sequence. Clearly, the difference of any two elements is another element of the sequence. Hence the sequence contains 0 so that the characteristic of F is nonzero. Theorem 2 The number of elements in a finite field F is p k , where p is a prime number. Proof: Let p be the characteristic of F . By the above, p > 0. As we know from the previous lecture, p is prime. Let F be the set of all elements 1 , 1+1 , 1+1+1 , . . . Clearly, F consists of p elements. One can show that F is a subfield (canonically identified with Z p ). It follows that F has p k elements, where k = dim F as a vector space over F .
Algebra over a field Definition. An algebra A over a field F (or F -algebra ) is a vector space over F with a multiplication which is a bilinear operation on A . That is, the product xy is both a linear function of x and a linear function of y . To be precise, the following axioms are to be satisfied: (A0) for all x , y A , the product xy is an element of A ; (A1) x ( y + z ) = xy + xz and ( y + z ) x = yx + zx for x , y , z A ; (A2) ( λ x ) y = λ ( xy ) = x ( λ y ) for all x , y A and λ F . An F -algebra is associative if the multiplication is associative. An associative algebra is both a vector space and a ring. An F -algebra A is a Lie algebra if the multiplication (usually denoted [ x , y ] and called Lie bracket in this case) satisfies: (Antisymmetry) : [ x , y ] = [ y , x ] for all x , y A ; (Jacobi's identity) : [[ x , y ] , z ] + [[ y , z ] , x ] + [[ z , x ] , y ] = 0 for all x , y , z A .
Examples of associative algebras: The space M n ( F ) of n × n matrices with entries in F . The space F [ X ] of polynomials p ( x ) = a 0 + a 1 X + · · · + a n X n with coefficients in F . The space of all functions f : S F on a set S taking values in a field F . Any field F that is an extension of a field F is an associative algebra over F . Examples of Lie algebras: R 3 with the cross product is a Lie algebra over R . Any associative algebra A with a Lie bracket (called the commutator ) defined by [ x , y ] = xy yx .