walters
  • walters
consider the groups Z3 and U(9) list the element of Z3 multiply U(9)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Z3 ={0,1,2} U(9) = {1,2,4,5,7,8}
anonymous
  • anonymous
now are u supposed to multiply Z3 by U(9)?
experimentX
  • experimentX
looks like both of you are on same school

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walters
  • walters
List the element of\[z _{3}\]multiply U(9)
walters
  • walters
yes i must multiply z3 by u(9)
experimentX
  • experimentX
what is Z3 and U9 ??
anonymous
  • anonymous
is it multiply or cross(X)
walters
  • walters
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
experimentX
  • experimentX
under what operation is it a group?
walters
  • walters
multiplication
experimentX
  • experimentX
how?? Z3 is not closed under multiplication
walters
  • walters
check question 3
walters
  • walters
i don't know whether tht x mean multiplication or cross
experimentX
  • experimentX
question no??
walters
  • walters
question 3 3.1 a
anonymous
  • anonymous
you should take an element from Z3 and combine it with an element from U9
walters
  • walters
"by combine wat do u mean "
experimentX
  • experimentX
LOL Q 5.4 came in my exam
walters
  • walters
z3={0,1,2} u(9)={1,2,4,5,7,8}
anonymous
  • anonymous
Take an the first element from z3 and combine it with all elements in u9
walters
  • walters
then z3 x u(9)={0,1,2,4,5,7,8}
anonymous
  • anonymous
u get (0,1) (0,2) (0,4) (0,5) (0,7) (0,8)
anonymous
  • anonymous
then u have to do the same with the second element of Z3
walters
  • walters
ok
walters
  • walters
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)
experimentX
  • experimentX
http://www.physicsforums.com/showthread.php?t=474271
experimentX
  • experimentX
looks like by group U(n), you mean \( \Bbb Z \mod n \) what is the convention followed for Z in your book?
walters
  • walters
but the way beketso did this it is not easy to get z3 cross u(9)
anonymous
  • anonymous
by U(n) we mean a set of numbers which are relatively prime to n and are smaller than n
experimentX
  • experimentX
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
@experimentX Z is the set of integers and Zn is Z mod n
experimentX
  • experimentX
what's the operation? it talks about cyclic group
walters
  • walters
i think since we r in z multiplication and addition are closed .
experimentX
  • experimentX
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?
walters
  • walters
First course in abstract algebra by fraleigh
walters
  • walters
i think it will be closed under *
anonymous
  • anonymous
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)
walters
  • walters
because by addition ther is a high possibility wen u add 2 integar u find the resul will no longer suits u(9)
anonymous
  • anonymous
on which operation is U9 a group?
walters
  • walters
addition
walters
  • walters
because by multiplication it won't hold ie 5*7 is not in u(9)
anonymous
  • anonymous
5*7 = 8 which is in U(9)
walters
  • walters
so it means the operation is either + or *
walters
  • walters
7+8 is also 8 wen working with addition
anonymous
  • anonymous
7+8 = 6 in U(9) which is not in the set
anonymous
  • anonymous
that is why i think the identity element is (0,1)
walters
  • walters
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)
walters
  • walters
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
experimentX
  • experimentX
your text book is very nice. the problem looks like it's related to Group Action on sets
walters
  • walters
because u r saying the identity is (0,1) so meaning the operation will be *
experimentX
  • experimentX
but it's morning here ... i gotta get some sleep. will see later
walters
  • walters
it is also morning to me see u
experimentX
  • experimentX
it's 7 am in the morning!!
walters
  • walters
to me is 3:12 am
anonymous
  • anonymous
@walters will do this more later
experimentX
  • experimentX
see ya later .. will see this tomorrow.

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