Exam 5 This review sheet seems much longer because I'm trying to provide some examples too. *Know the definiton of a Binary Operation: A Binary Operation on a set S is a symbol which associates to each pair of elements from S a new element from S EXAMPLE Let S = {a, b, c} then the operation * defined by the following table is a binary operation: * a b c +---------------------- a | a a b b | b c b c | c b a This table says that a*b = a while b*a = b , so this operation is NOT commutative Also this table says that a*(b*c) = a*b = a while (a*b)*c = a*c = b, so this operation is NOT associative. *Know the definition for a Group, a Ring, and a Field A Group is a set with a binary operation + that satisfies the following algebraic laws: 1) Associativity -- a + (b + c) = (a + b) + c 2) Identity -- There is an identity element 0 such that for each x 0 + x = x + 0 = x 3) Inverses -- Each element x has an inverse -x such that x + (-x) = (-x) + x = 0 A Ring adds a second binary operation * and the following algebraic laws to the ones for a group: 4) + is commutative -- a + b = b + a 5) * is associative -- a*(b*c) = (a*b)*c 6) * has an identity -- There is a multiplicative identity 1 such that 1*x = x*1 = x for each x D) Distributivity -- a*(b + c) = a*b + a*c AND (b + c)*a = b*a + c*a A Field adds two more algebraic laws to the ones for a ring: 7) * has (multiplicative) inverses for each element 8) * is commutative REMARKS: (1) For rings and fields we always call the operations + and *. For groups we sometimes call the operation + (plus) and sometimes * (times) depending on the situation. (2) A Group is Abelian if it also satisfies the 4th law above, that is, if the single group operation happens to be commutative. *Know the examples we gave for Groups: There are some "familiar" examples which are pretty similar to each other: ----- the integers with (familiar) addition ----- the rational numbers with (familiar) addition ----- the real numbers with (familiar) addition If we let Q* = Q - {0} and R* = R - {0}, that is, we throw out the number zero then we have groups: ----- the rational numbers without zero under (familiar) multiplication ----- the real numbers without zero under (familiar) multiplication If you read the symbol Z_n as "Z sub n" or "The integers modulo n" then ---- the integers modulo n under addition modulo n We also talked about the following NON-abelian groups: If _Q_ means boldface Q then _Q_ = Quaternions = {1, -1, i, -i, j, -j, k, -k} and ----- Hamilton's Quaternion group ---- The symmetric group on three objects *Know how to build a group table (that is, do the group operations) for _Q_, S_3, and the various Z_n EXAMPLE The group table for Q is as follows: * 1 -1 i -i j -j k -k +------------------------------ 1| 1 -1 i -i j -j k -k -1|-1 1 -i i -j j -k k i| i -i -1 1 k -k -j j -i|-i i 1 -1 -k k j -j j| j -j -k k -1 1 i -i -j|-j j k -k 1 -1 -i i k| k -k j -j -i i -1 1 -k|-k k -j j i -i 1 -1 REMARK: the "circle diagram" helps to calculate here: i --> j --> k --> i going against the arrow adds a minus sign and i^2 = j^2 = k^2 = -1 EXAMPLE Here are some examples from S_3 (imagine that there are big parentheses on the left and the right) (1 2 3) * (1 2 3) = (1 2 3) (2 1 3) (3 2 1) (3 1 2) (1 2 3) * (1 2 3) = (1 2 3) (3 1 2) (3 1 2) (2 3 1) *Know how to use these tables find the orders of elements in various groups. Remember the order of an element is the least number of times you have to add/multiply that element with itself to get the identity of that group. *Know how to identify isomorphic groups from their tables. *Know how to reduce large powers of a number modulo another number EXAMPLE 1 Reduce 3^101 (mod 4) Step 1: Make a table of the powers of 3 and reduce them mod 4 until you get 1 3^1 = 3 (mod 4) 3^2 = 9 = 1 (mod 4) Step 2: Divide 101 by this power and let R equal the remainder 101 divided by 2 is 50 remainder 1, so R = 1 Step 3: The answer is 3^R reduced mod 4 3^101 = 3^1 = 3 (mod 4) EXAMPLE 2 Reduce 7^854 (mod 9) Step 1: Make a table of the powers of 7 and reduce them mod 9 until you get 1 7^1 = 7 (mod 9) 7^2 = 49 = 4 (mod 9) 7^3 = 4 * 7 = 28 = 1 (mod 9) Step 2: Divide 854 by this power and let R equal the remainder 854 divided by 3 is 284 remainder 2, so R = 2 Step 3: The answer is 7^R reduced mod 9 7^854 = 7^2 = 4 (mod 9) *Theorems +In a Group the Identity element is unique +In a Group the Inverse of an element is unique +In a Ring 0*x = 0 for each x +In a Ring (-1)*x = -x for each x *For Group Theory know that in the group table each element appears exactly once in each row and each column.