Let $p$ be a prime number which doesn't divide $2mnr$. So $p$ is a unit in the ring $\mathbb{Z}/2mnr \mathbb{Z}$ and $q=p^k$ for a certain $k \in \mathbb{Z}$
Could you explain to me why then:
1) $2mnr$ divides $q-1$ and
2) in the group $\mathbb{F}_q ^{\times}$ which has order $q-1$ there exist elements $a, \ b, \ c$ which have orders $2m, \ 2n, \ 2r$ respectively.
I would really appreciate all your help.
Thank you.
Both claims are false.
For example, let $m=5$, $n=7$, $r=11$, $p=13$, and $k=1$. Then $$2mnr=2\cdot 5\cdot 7\cdot 11=770$$ doesn't divide $q-1=12$, and $\mathbb{F}_q^\times\cong\mathbb{Z}/12\mathbb{Z}$ has no elements of order $2m=10$, $2n=14$, or $2r=22$.
Presumably $\rm\ 2mnr\mid q\!-\!1 = p^k\!-1\ $ is true by hypothesis, e.g. if $\rm\:k\:$ is the order of $\rm\:p\,\ (mod\ 2mnr).\:$ Then, since the multiplicatiove group of a finite field is cyclic, $\rm\:G = \mathbb{F}_q ^{\times}\:$ is cyclic of order $\rm\:q\!-\!1\:$ divisible by $\rm\:2m,2n,2r,\:$ elements exist of those orders, e.g. $\rm\:G = \langle g\rangle\:\Rightarrow\:g^{(q-1)/(2n)}$ has order $\rm\:2n.$
Edit $\ $ The surmisal is correct, as confirmed by the link just posted (should be p.29).
