Proof: Suppose that d(1) , d(2) , ...., d(k) are all the elements of order d.
Look at the subgroups <d(1) >, <d(2) >, ... <d(k) >
(i) Each of these is cyclic and so each has phi(d) elements of order d.
(ii) Suppose that some element x of order d is in two of these subgroups
x in <d(i)> intersection <d(j)>. Then <d(i)> = <x> = <d(j)>
(iii) We count the number of these subgroups.
no two can share any elements of order d.
The number of elements of order d will be:
(the number of distinct subgroups) * phi(d).
Can somebody explain me in part (ii) why <d(i)> = <x> = <d(j)> ?