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