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.

Algebra 101

Why would I want to though?

In a cyclic group, order n, there is one and only one subgroup, order f, if f is a divisor of f.
(x^(n/f)) is the one you need.

Why is it the only one ? If (x^k) is another groupe of order f, then you have x^kf= e. So the order of G is a divisor of kf and then n/f is a divisor of k. But then x^k is in (x^(n/f)).
So you have (x^k) include in (x^(n/f)) and because of their equal order, (x^k)=(x^(n/f)).

So there is a unique subgroup of order f, name it H, in your fucking cyclic group (which, btw, is Z/nZ). The generators of H are all the integers a that verify : gcd(a,d) = 1 (dunno how you call it in english). They are all of order d,no one other is of order d and they are counted by the fucking euler phi function.

There you go, fgsfds

>>4
I still don't really get it but thanks.

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.

>>6

Your english is fine for talking about maths. Any non-retarded person should understand you.