# Math Help - In a finite group, the number of elements of order d is divisible by phi(d)

1. ## In a finite group, the number of elements of order d is divisible by phi(d)

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)> ?

2. If $x \in \cap$, then $ \subset \cap$. But can this containment be proper, if we're given that $||=d$?

3. Thank you, I got it. if they are finite set and one is contained in the other, then they must be equal set. Now it makes sense.