Question

group theory: a has order n, m is a factor of n, C_m := {x in | x^m = e} prove an element x in has order m iff x is a generator of C_m

Answer 1

this has something to do with a gcd i believe

Answer 2

assume gcd(m,n) not= 1 ....

Answer 3

I know from the problem above that C_m has m elements

Answer 4

i recall something about an example of mod 6

<0>: 0,1,2,3,4,5
<2>: 0,2,4
<3>: 0,3

Answer 5

that may have been better as <1> instead of <0> :)

Answer 6

ahh:)

Answer 7

so 2 and 4 have order 2 on mod 6?

Answer 8

i cant recollect the process that clearly, but something about the order is a gcd not equal to 1

yes

Answer 9

0,4,8,12
0,4,2,0 ...

Answer 10

order 3

Answer 11

yes 3, of well at least the question is starting to make sense to me:)

Answer 12

<5> = 0,5,15,20,25,30
0,5,3,2,1

hmm, since gcd(5,6) = 1, im missing a 4 spot

Answer 13

0 5 10
0 6 6+4

lol

Answer 14

its like guass elem, the mistakes are the easy part

Answer 15

:)

Answer 16

there is also a "grid" pattern of some effect, not proof worthy of course

Answer 17

yeah but it helps

Answer 18

mod 9, operator grid can be divvied up into
1 1 1 1 1 1 1 1 1 1
3 3 3

a mod 8 can be divvied up as:
1 1 1 1 1 1 1 1
2 2 2 2
4 4
8

mod 6
1 1 1 1 1 1
2 2 2
3 3
6

Answer 19

what are the rows saying?

Answer 20

just recalling the different divisions of the rows and columns into sungroups

Answer 21

1 and 3 only generate mod 9?