Finding the other elements of order 8

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?

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.

Answer 1

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 $n$.

Answer 2

Question

Suppose that $G$ is a cyclic group and that $6$ divides the order of $G$.

1. How many elements of order $6$ does $G$ have?
2. If $8$ divides order of $G$, how many elements of order $8$ does $G$ have?
3. 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 $\Bbb Z_n$.

1. 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.

2. Rinse and repeat to obtain $8$.

3. Since the subgroup of order $8$ is unique, the elements are $a$, $a^3$, $a^5$, and $a^7$.