Proof
The first statement implies the second.
The second statement implies the first.
The second statement implies the third.
The third statement implies the second.
G can be identified with Z/nZ. Let
d be a divisor of n. An element x of
Z/nZ has order d in the additive group of
Z/nZ if and only if gcd(x,n) =
n/d. Dividing both x and n by n/d
(their gcd) we see that the number of such x is equal to 
(d).
(d).
Since (n) > 0 there is an element g
of order n in G. Hence G is cyclic.
(n) > 0 there is an element g
of order n in G. Hence G is cyclic.
This follows from the
formula
n = 
d|n,d>0(d).
Indeed, every element x with xd = e
has order dividing d. By the above formula, the sum of the
number of elements of order m over all divisors m of
d, equals d.
d|n,d>0(d).
Indeed, every element x with xd = e
has order dividing d. By the above formula, the sum of the
number of elements of order m over all divisors m of
d, equals d.
Let 
(d) denote the number of elements
in G of order d. Clearly (1) =
1. For d > 1 we have the formula d = m|d,m>0(m), since we can divide the set of elements
x with xd = e into disjoint parts
consisting of those elements that have order m, where m
runs through the divisors of d. This implies that and satisfy the same
recursive formula, and thus are the same. In particular we have
proved the second statement.
(d) denote the number of elements
in G of order d. Clearly (1) =
1. For d > 1 we have the formula d = m|d,m>0(m), since we can divide the set of elements
x with xd = e into disjoint parts
consisting of those elements that have order m, where m
runs through the divisors of d. This implies that and satisfy the same
recursive formula, and thus are the same. In particular we have
proved the second statement.