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anonymous
 3 years ago
consider the groups Z3 and U(9)
list the element of Z3 multiply U(9)
anonymous
 3 years ago
consider the groups Z3 and U(9) list the element of Z3 multiply U(9)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Z3 ={0,1,2} U(9) = {1,2,4,5,7,8}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now are u supposed to multiply Z3 by U(9)?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0looks like both of you are on same school

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0List the element of\[z _{3}\]multiply U(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i must multiply z3 by u(9)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0what is Z3 and U9 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it multiply or cross(X)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0z3 set of integers and u(9) means u find the number from 1 to 9 that cannot divide 9 or and those number must never divide each other

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0under what operation is it a group?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0how?? Z3 is not closed under multiplication

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i don't know whether tht x mean multiplication or cross

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you should take an element from Z3 and combine it with an element from U9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"by combine wat do u mean "

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0LOL Q 5.4 came in my exam

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0z3={0,1,2} u(9)={1,2,4,5,7,8}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Take an the first element from z3 and combine it with all elements in u9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then z3 x u(9)={0,1,2,4,5,7,8}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u get (0,1) (0,2) (0,4) (0,5) (0,7) (0,8)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then u have to do the same with the second element of Z3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh becuase u came with a new set how will u get the identity because the new set is different to z3 and u(9)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0looks like by group U(n), you mean \( \Bbb Z \mod n \) what is the convention followed for Z in your book?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the way beketso did this it is not easy to get z3 cross u(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by U(n) we mean a set of numbers which are relatively prime to n and are smaller than n

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0you sure on that?? (how is it closed on operation if it has no mechanism to reduce elements greater than it) and what is Z?? check your book again ... what convention is followed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0z3 x U9 ={(0,1) (0,2) (0,4) (0,5) (0,7) (0,8) (1,1) (1,2) (1,4) (1,5) (1,7) (1,8) (2,1) (2,2) (2,4) (2,5) (2,6) (2,8) } The first element is from z3 and the second is from U(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX Z is the set of integers and Zn is Z mod n

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0what's the operation? it talks about cyclic group

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think since we r in z multiplication and addition are closed .

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0since you have taken Cartesian product, \( \Bbb Z_3 \times U(9) \) how do you define this new group? What book are you following for abstract algebra?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First course in abstract algebra by fraleigh

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think it will be closed under *

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0am not sure but i think it is the combination of the identity element from Z3 and the identity element from U9 So, the identity element is (0,1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because by addition ther is a high possibility wen u add 2 integar u find the resul will no longer suits u(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on which operation is U9 a group?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because by multiplication it won't hold ie 5*7 is not in u(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05*7 = 8 which is in U(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it means the operation is either + or *

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.07+8 is also 8 wen working with addition

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.07+8 = 6 in U(9) which is not in the set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is why i think the identity element is (0,1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it can't be the values will circulate in {1,2,4,5,7,8} only there is no way we can get any value except those who are in u(9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we can use the fact that says most of the time wen the first elementin a set is 0 in the the the operation that likely to be closed is + and if the fist element is 1 * is likely to be closed

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0your text book is very nice. the problem looks like it's related to Group Action on sets

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because u r saying the identity is (0,1) so meaning the operation will be *

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0but it's morning here ... i gotta get some sleep. will see later

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is also morning to me see u

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0it's 7 am in the morning!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@walters will do this more later

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0see ya later .. will see this tomorrow.
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