Hail Mathematicians,
Problem:
Prove that iff
Stuff that might come in handy:
Perhaps some notation will be of assistance.
Let be a group and . We denote the subgroup generated by as
For anything else, Ask, and it shall be given unto you.
What I tried:
( ). Assume . Then we have (this is right, right?)
By this problem:
Say is odd. Then we have for some .
and i don't know where to go from there.
Say is even. Then for some .
and i don't know where to go from there.
I do know that i want to show that for . Furthermore, I have to make sure 1 is the gcd by showing that 1|a, 1|n and if there is some b such that b|a and b|n then b|1.
( ) For the converse, assume . Then for some .
and i don't know where to go from there.
Help
Thanks guys and gals