Finding elements of order $8$ in $\mathbb{Z}_{8000000}$. I want to find elements of order $8$ in $\mathbb{Z}_{8000000}$. 
I know that elements of order $n$ in a group $\mathbb{Z}_m$ is $\phi(n)$. But how can I apply this result for a group having such a large number of elements?
Thanks for the help.
 A: Since $Z_{8000000}$ is cyclic, it contains exactly $\phi(8)=4$ elements of order $8$. We see that $1000000\in Z_{8000000}$ is one of these elements. It is also the generator of the unique subgroup of $Z_{8000000}$ of order $8$. The other three elements of order $8$ are $7000000$, $5000000$, and $3000000$. In general, the generators of this unique subgroup are the elements $n\cdot
 1000000$, where $\gcd(n,8)=1$.
NOTE: $\phi(8000000)$ would be the number of generators of $Z_{8000000}$. We calculate $\phi(8)$ because we are interested in the number of generators of the subgroup of $Z_{8000000}$ of size $8$ as opposed to the whole group, which is size $8000000$.
A: The order of $\bar m$ in $\mathbb{Z}_{8000000}$ is $\frac{8000000}{gcd(m,8000000)}$. This is $8$ if and only if 
$$gcd(m,8000000)=1000000$$
Thus, $m$ must be an odd multiple of $1000000$.
A: There are $4$ elements of order $8$. This is because Φ(8)=4, and these elements are $1,3,5,7$. So the generators of $8000000$ of order $8$ are $1000000,3000000,5000000$ and $7000000$.
