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Abstract Algebra




                    Notes          Now (a * a) * b = a * b = e.
                                   Also a * (a * b) = a * e = a. Therefore, by Gl’, a = e.

                                   Now we will use this lemma to prove Theorem 3.
                                   Proof of Theorem 3: G1 holds since G1 and G1' are the same axiom. We will next prove that G3
                                   is true. Let a  G. By G3' 3 b  G such that a * b = e. We will show that

                                   b * a = e. Now,
                                   (b * a) * (b * a) = (b * (a * b)) * a = (b * e) * a = b * a .
                                   Therefore, by Lemma 1, b * a = e. Therefore, G3 is true.
                                   Now we will show that G2 holds. Let a  G. Then by G2', for a  G, a * e = a. Since G3 holds,
                                    b  G such that a * b = b * a = e. Then
                                   e * a = (a * b) * a = a * (b * a) = a * e = a .
                                   That is, G2 also holds.

                                   Thus, (G, *) satisfies G1, G2 and G3.

                                         Example:  Let G = { ±1,  i }, i =    Let the binary operation be multiplication. Show
                                                                     1.
                                   that (G) is a group.
                                   Solution: The table of the operation is

                                                         1              –1              i             –i
                                          1              1              –1              i             –i
                                          –1             –1             1              –i              i
                                          i              i              –i             –1              1
                                          –i             –i              i             1              –1

                                   This table shows us that a.l = a   a  G. Therefore, 1 is the identity element. It also shows us that
                                   (G) satisfies G3. Therefore, (G) is a group.
                                   Note that G = {1, x, x , x }, where x = i.
                                                      3
                                                   2
                                   2.2.1 Abelian  Group

                                   Definition: If (G, *) is a group, where G is a finite set consisting of n elements, then we say that
                                   (G, *) is a Finite group of order n. If G is an infinite set, then we say that (G,*) is an infinite group.

                                   If * is a commutative binary operation we say that (G, *) is a commutative group, or an abelian
                                   group. Abelian groups are named after the gifted young Norwegian mathematician Niels Henrik
                                   Abel.
                                   Now let us discuss an example of a non-commutative (or non-abelian) group. Before doing this
                                   example recall that an m x n matrix over a Set S is a rectangular arrangement of elements of S in
                                   m rows and n columns.


                                                   a b 
                                            If A =      then ad-bc is called the determinant of A and is written as det
                                      Note         c d
                                     A or |A|





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