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

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

looks like both of you are on same school

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

yes i must multiply z3 by u(9)

what is Z3 and U9 ??

is it multiply or cross(X)

under what operation is it a group?

multiplication

how?? Z3 is not closed under multiplication

check question 3

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

question no??

question 3
3.1 a

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

"by combine wat do u mean "

LOL Q 5.4 came in my exam

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

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

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

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

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

ok

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

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

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

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

what's the operation? it talks about cyclic group

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

First course in abstract algebra by fraleigh

i think it will be closed under *

on which operation is U9 a group?

addition

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

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

so it means the operation is either + or *

7+8 is also 8 wen working with addition

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

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

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

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

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

it is also morning to me see u

it's 7 am in the morning!!

to me is 3:12 am

see ya later .. will see this tomorrow.