Products Of Primitive Roots Of Unity

Relevant wiki: Primitive Roots of Unity

Let $\large \zeta_{m}=e^{\frac{2\pi k_1i}{m}},\zeta_{n}=e^{\frac{2\pi k_2i}{n}},\zeta_{l}=e^{\frac{2\pi k_3i}{l}}$

As per conditions we have $\zeta_{m}\zeta_{n}=\zeta_{l}\implies \frac{k_1}{m}+\frac{k_2}{n}=\frac{k_3}{l}$ .

Moreover Since it's a $\text{primitive root of unity}$ , We have $(k_1,m)=(k_2,n)=(k_3,l)=1$ where $k_1\in[1,m],k_2\in[1,n],k_3\in[1,l]$

$\text{Case 1:}$ Suppose $l=gcd(m,n)\implies gcd(\frac{m}{l},\frac{n}{l}=1)$ , $(k_1,m)=1\implies (k_1,\frac{m}{l})=1$ . Similarly $(k_2,\frac{n}{l})=1$ .

So $\large \frac{k_1}{\frac{m}{l}} + \frac{k_2}{\frac{n}{l}} = k_3$ , RHS is an integer but LHS is irreducible since $(k_1,\frac{m}{l})=1,(k_2,\frac{n}{l})=1$ . $\text{Case Dismissed}$

$\text{Case 2:}$ Suppose $l=mn$ . So we have the following facts :

$\begin{cases} (k_3,l)=1 \\ (k_3,mn)=1 \\ (k_3,m)=1 \\(k_3,n)=1\end{cases}$

Putting $l=mn$ we get , $\frac{k_1n}{k_3}+\frac{k_2m}{k_3}=1$ . So $k_3|k_1,k_2$ which is a special requirement for this identity to be satisfied and not always true. So $\text{Case Dismissed}$

$\text{Case 3 :}$ Suppose $l=lcm(m,n), l=mx=ny$ with $(x,y)=1$ , $(k_3,mx)=(k_3,ny)=1\implies (k_3,m)=(k_3,x)=(k_3,n)=(k_3,y)=1$ . So we get putting $m=\frac{l}{x},n=\frac{l}{y}$ ,

$k_1x+k_2y=k_3\implies k_1\frac{x}{k_3} + k_2\frac{y}{k_3}=1$ . Again since RHS is an integer we must have LHS integer.

So, $k_3|k_1,k_2$ which is the requirement for $l=lcm(m,n)$ to be true. So in general we can't say this is always true. $\text{Case Dismissed}$

$\text{So none of the options fit so we have, None Of These}$

An even easier counterexample comes from $m=n=2$ and $\ell=1$ .