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  Abstract Algebra Anonymous (For Reddit)Spring 2014 Introduction The purpose of this text is nothing but personal recreation. At times, I feelthat the best way to improve one’s understanding of a subject is to throwaway the books and exercises, and write down what you remember in amanner that fits your personal intuition. If I have decided to share this withyou, I cannot guarantee that you will benefit from reading this text at anylevel. I do not take responsibility for any errors or inaccuracies in the text.While I do intend to try to write defintions, theorems and proofs in my ownwords, I may have trouble with this, especially if I am unfamiliar with thetopic. Therefore, some parts of the text may be based on  A first course in Abstract Algebra   by Fraleigh. 1 Sets, Binary Operations and Groups A  set   is a collection of objects. Naturally, we are mostly working with num-bers, but  { ! , @ , ? }  is still a set, by definition. We may construct a set asfollows:  { 2 n | n  ∈  Z }  to create the set of all even numbers. An informal defi-nition of a  binary operation   is an operation on a set, that when performed ontwo members of the set returns another member of the set. As such, addi-tion is not a binary operation on the set  { 1 , 2 , 3 } , because 2+3 = 5, which isnot in the set. However, addition is a binary operation on the set  { n | n  ∈  Z } .A  group  is a set under a binary operation that meets the criteria for abinary operation above, from now on called  closure  . In addition, any groupmust have an identity element  e , with the property that  a  ∗  e  =  e  ∗  a  =  a where  ∗  is our binary operation and  a  is any element in our group. Further-more, for every element  a  ∈  G , where  G  is our group, there must exist aninverse element  a  with the property that  a ∗ a  =  a  ∗ a  =  e . Our last criterion1  is associativity: for any  a,b,c  ∈  G  we have that  a ∗ ( b ∗ c ) = ( a ∗ b ) ∗ c  =  a ∗ b ∗ c .One can play around with examples to get a better grip on the definition.For example, I claim that  < Z , + >  is a group, while  < Z , · >  is not.A  subgroup  H   of a group  G  needs to meet all the axioms above. In ad-dition, all elements in  H   must be in  G . It follows that  H   and  G  must sharethe identity element and that  H   is closed under the operation and  G , thatis, for all  a,b  ∈  H   we have  a  ∗  b  ∈  H   and thus  a  ∗  b  ∈  G . 1.1 Cyclic Groups A group  G  is said to be  cyclic   if there is an element  a  ∈  G  that  generates  G , that is,  < { a,a 2 ,a 3 ... } , ∗ >  =  G , where  a 2 =  a  ∗  a . An element with thisproperty is called a  generator   of   G . While I have, for simplicity, writtenpositive exponents, this also works the other way around (with inverses),and being consciously aware of this may save you from an embarassingmisunderstanding, as showed in this example: Example 1.1.  Determine if   G  =  < Z , + >  is cyclic.Note that, given our initial definition of a cyclic group, where we disregardedinverses,  G  would not be cyclic. However, I claim that 1 is a generator for  G .Note that  − 1 is the inverse of 1, and we have  { ... − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 ... }  =  Z as a result of adding both the inverse and the element itself to 1. Thus,  G is cyclic. A personal anecdote is me answering that  < Q , + >  is not cyclicbecause it is infinite, which is the right answer with the wrong reason. Thisbrings us to our next example: Example 1.2.  Determine if   G  =  < Q , + >  is cyclic.No, and here is the  right   reason: on every interval [ a,b ] there are infinitelymany irrational and rational numbers. Therefore there may not be an ele-ment  a  ∈  G  that generates the whole group, seeing as some rational numbers(infinitely many, in fact) are skipped in the  a  →  2 a -process.We often see cyclic groups on the form  Z n , which can be read as  Z  mod-ulo  n , where  a  modulo  b  means ”the remainder of   a  when divided by  b ”.Therefore, 1 + 2 = 0 in  Z 3 , as 3 modulo 3 = 0.2  Theorem 1.1.  Every element in a cyclic group  G  generates a subgroup of  G . Furthermore, this subgroup is also cyclic. Proof: 3
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