Question

In a finite group ,show that the number of elements of order d is a.multiple of phi(d)

Answer 1

Solution:

We will use the following theorem in our proof.

Theorem 1: Number of Elements of Each Order of a Cyclic Group

Proof:

Let G be a finite group. If G has no elements of order d, $\phi(d) \mid 0$ . So suppose $a \in G$ with |a|=d. By Theorem 1, $\langle a\rangle\ had\ \phi(d)$ elements of order d. If all elements of order d are in $\langle a\rangle$ , we are done. So suppose $b \in G,|b|=d, b \notin\langle a\rangle$ . Then $\langle b\rangle$ also has $\phi(d)$ elements of order d. Thus, if $\langle a\rangle$ and $\langle b\rangle$ have no elements of order d in common, we have found 2 $\phi(d)$ elements of order d. If |c|=d and $c \in\langle a\rangle \bigcap\langle b\rangle,\langle a\rangle=\langle c\rangle=\langle b\rangle,$ a contradiction.

Continuing, the number of elements of order d is a multiple of $\phi(d)$.

Hence, proved.