Originally Posted by

**Benmath** Hi,

So I have a problem that I have been looking at for 4 days now.

Let F be the quotient group G/Z (Z is integers) where G= { ap^b in Q| a, b in Z} where Q is the rational numbers and Z is the integers.

Let let x,y be in F and o(x)=< o(y) (order of x is less or equal to order of y) if and only if x is in <y> (cyclic group generated by y).

Going <== (backwards direction) is fine. I completely understand this direction. Going ==> is where I'm having the trouble.

==> Let x,y in F. Then x=ap^b + Z and y= cp^d + Z. I know the o(x)=< o(y). My professor says that by re-writing ap^b (because it is rational), I can find the exact order of what x is (the smallest number I can multiply x by to get an integer). I tried using actual numbers, but I still cannot get what he was talking about. Also, even if i get what the exact order of x is, I have no idea how to use that to get that x is in <y>.

I cannot seem to understand how using the order will tell me that there exists so t in Z such that x=ty. (which would put x in <y>).

Any help will be great. Thank you.