you should be able to solve this

G = (x) is a cyclic group, order n.
f is a divisor of n.
prove there are φ(f) elements of order f in G, where φ is the euler phi function.

The fucking goddamn euler phi function count the numbers a that are between 1 and f and verifies : gcd(a,f) = 1.
They also are the generators of a cyclic group of order f. So their order are f, for all, ie the euler fag function counts the generators of a cyclic group of order f.
I demonstrate that there is a unique subgroup H of order f in G. SO if you consider an element of order f, it is forcefully in H (the fucking Lagrange theorem).
SO EVERY FUCKING ELEMENT OF ORDER F IS COUNTED BY PHI FGSFDS.

Protip : In fact, there is one cyclic group of order n (isomorphicly speaking, fuck this retarded english language, I can't even know if you will understand this). It's Z/nZ. The cyclics subgroups, if f is a divisor of n, are like Z/fZ. If you have problems to see my point when considering an abstract G cyclic group, then think about  this.