Question

consider the groups Z3 and U(9)
list the element of Z3 multiply U(9)

Answer 1

Z3 ={0,1,2}
U(9) = {1,2,4,5,7,8}

Answer 2

now are u supposed to multiply Z3 by U(9)?

Answer 3

looks like both of you are on same school

Answer 4

List the element of\[z _{3}\]multiply U(9)

Answer 5

yes i must multiply z3 by u(9)

Answer 6

what is Z3 and U9 ??

Answer 7

is it multiply or cross(X)

Answer 8

z3 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

Answer 9

under what operation is it a group?

Answer 10

multiplication

Answer 11

how?? Z3 is not closed under multiplication

Answer 12

check question 3

Answer 13

i don't know whether tht x mean multiplication or cross

Answer 14

question no??

Answer 15

question 3
3.1 a

Answer 16

you should take an element from Z3 and combine it with an element from U9

Answer 17

"by combine wat do u mean "

Answer 18

LOL Q 5.4 came in my exam

Answer 19

z3={0,1,2} u(9)={1,2,4,5,7,8}

Answer 20

Take an the first element from z3 and combine it with all elements in u9

Answer 21

then z3 x u(9)={0,1,2,4,5,7,8}

Answer 22

u get
(0,1) (0,2) (0,4) (0,5) (0,7) (0,8)

Answer 23

then u have to do the same with the second element of Z3

Answer 24

ok

Answer 25

oh becuase u came with a new set how will u get the identity because the new set is different to z3 and u(9)

Answer 26

http://www.physicsforums.com/showthread.php?t=474271

Answer 27

looks like by group U(n), you mean \( \Bbb Z \mod n \)
what is the convention followed for Z in your book?

Answer 28

but the way beketso did this it is not easy to get z3 cross u(9)

Answer 29

by U(n) we mean a set of numbers which are relatively prime to n and are smaller than n

Answer 30

you 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

Answer 31

z3 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)

Answer 32

@experimentX Z is the set of integers and Zn is Z mod n

Answer 33

what's the operation? it talks about cyclic group

Answer 34

i think since we r in z multiplication and addition are closed .

Answer 35

since 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?

Answer 36

First course in abstract algebra by fraleigh

Answer 37

i think it will be closed under *

Answer 38

am 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)

Answer 39

because by addition ther is a high possibility wen u add 2 integar u find the resul will no longer suits u(9)

Answer 40

on which operation is U9 a group?

Answer 41

addition

Answer 42

because by multiplication it won't hold ie 5*7 is not in u(9)

Answer 43

5*7 = 8 which is in U(9)

Answer 44

so it means the operation is either + or *

Answer 45

7+8 is also 8 wen working with addition

Answer 46

7+8 = 6 in U(9) which is not in the set

Answer 47

that is why i think the identity element is (0,1)

Answer 48

it 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)

Answer 49

we 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

Answer 50

your text book is very nice. the problem looks like it's related to Group Action on sets

Answer 51

because u r saying the identity is (0,1) so meaning the operation will be *

Answer 52

but it's morning here ... i gotta get some sleep. will see later

Answer 53

it is also morning to me see u

Answer 54

it's 7 am in the morning!!

Answer 55

to me is 3:12 am

Answer 56

@walters will do this more later

Answer 57

see ya later .. will see this tomorrow.