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walters
consider the groups Z3 and U(9) list the element of Z3 multiply U(9)
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)
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
under what operation is it a group?
how?? Z3 is not closed under multiplication
i don't know whether tht x mean multiplication or cross
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
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)
looks like by group U(n), you mean \( \Bbb Z \mod n \) what is the convention followed for Z in your book?
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
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
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)
@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 .
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?
First course in abstract algebra by fraleigh
i think it will be closed under *
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)
because by addition ther is a high possibility wen u add 2 integar u find the resul will no longer suits u(9)
on which operation is U9 a group?
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)
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)
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
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!!
@walters will do this more later
see ya later .. will see this tomorrow.