Finding the other elements of order 8 Suppose that G is a cyclic group and that 6 divides the order of G. How many elements of order 6 does G have? If 8 divides order of G, how many elements of order 8 does G have? If a is an element of order 8, what are the other elements of order 8?
Using the Euler phi totient:
There are 2 elements of order 6.
There are 4 elements of order 8.
If a is an element of order 8, that leaves 3 other elements of order 8.
The order of G is clearly 24 by the least common multiple of 8 and 6. 
I used the fact that for any positive integer k, the order of an element $a^{k}$ is n/gcd(n,k). This gives $a^{3}$ as another element of order 8. But this is where I cannot further my attempt.
Any hint is appreciated.
Thanks in advance. 
 A: Question
Suppose that $G$ is a cyclic group and that $6$ divides the order of $G$.


*

*How many elements of order $6$ does $G$ have?

*If $8$ divides order of $G$, how many elements of order $8$ does $G$ have?

*If $a$ is an element of order $8$, what are the other elements of order $8$?


Answer
A cyclic group of order $n$ is isomorphic to $\Bbb Z_n$.


*

*If $n=6k$ for some positive integer $k$, then finding an element with order $6$ is equivalent to solving $6x \equiv 0 \pmod{6k}$, which has exactly $6$ solutions ($0$, $k$, $2k$, $3k$, $4k$, and $5k$), so they form $1$ unique subgroup.

*Rinse and repeat to obtain $8$.

*Since the subgroup of order $8$ is unique, the elements are $a$, $a^3$, $a^5$, and $a^7$.
A: A cyclic group of order $n$ has exactly one subgroup of order $d$ for each $d$ dividing $n$.
This subgroup is cyclic and has $\phi(d)$ elements of order $d$.
The elements of order $d$ are $h^i$ for $\gcd(i,d)=1$, where $h$ is an element of order $d$.
All this does not depend on $n$.
