I’m confused how we concluded the unit group is cyclic. We showed there can’t be more than k elements with order dividing k and we know every element of the unit group divides k so we should get the inequality |F^x|<=k.
I also don’t understand how stating k=p-1 would imply that F^x is cyclic. This seems to be assuming that |F|=p, but in general a finite field can have size p^n and we would then have k=p^n-1. But I don’t understand the equality anyway and I don’t understand why that equality would say anything about F^x being cyclic?
I shouldn't have written p, but rather q, where q is the number of elements in F. In any case, I reworded it in a way that makes more sense. Remember: we know that there is an element with order k... but we've shown that k cannot be any smaller than the total number of elements in the unit group!
I’m confused how we concluded the unit group is cyclic. We showed there can’t be more than k elements with order dividing k and we know every element of the unit group divides k so we should get the inequality |F^x|<=k.
I also don’t understand how stating k=p-1 would imply that F^x is cyclic. This seems to be assuming that |F|=p, but in general a finite field can have size p^n and we would then have k=p^n-1. But I don’t understand the equality anyway and I don’t understand why that equality would say anything about F^x being cyclic?
I shouldn't have written p, but rather q, where q is the number of elements in F. In any case, I reworded it in a way that makes more sense. Remember: we know that there is an element with order k... but we've shown that k cannot be any smaller than the total number of elements in the unit group!
That fully clears things up, thank you!