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

• Sep 30th 2010, 09:17 AM
santiagos11
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)> ?
• Sep 30th 2010, 09:31 AM
Bruno J.
If $\displaystyle x \in <d_i>\cap<d_j>$, then $\displaystyle <x> \subset <d_i>\cap<d_j>$. But can this containment be proper, if we're given that $\displaystyle |<x>|=d$?
• Sep 30th 2010, 12:57 PM
santiagos11
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.