Proof that multiplication of primitive roots of unity is a primitive root of unity. I can't solve this exercise. Could you help me?
Let $w$ be a 38-th primitive root of unity. Let $z$ be a 22-nd primitive root of unity. Prove that ($w.z$) is a 209-th primitive root of unity.
What I have so far:
I understand that, in order to prove that ($w.z$) is a 209-th root of unity, I must prove that:
$(w.z)^k = 1 \iff k \equiv 0$ (209) 
I tried to brake down the "$\iff$" in two part.
First, the "$\Longleftarrow$" part:
I know that $k = 209 * q = 19*11*q$ (with q $\in \mathbb{N})$.
So $(w.z)^k$ = $w^{209}*z^{209}$ = $w^{19*11*q}*z^{19*11*j}$ = 
$(w^{19})^{11*q}.(z^{11})^{19*j}$ = $1^{11*q}1^{19*j}$ = $1$ (Because w is a 38=19*2-th root of unity and z is a 22=11*2-th root of unity)
Is this correct? 
Also, I don't know how to prove the "$\Longrightarrow$" implication.
Thanks!
 A: For the '$\Leftarrow$' part, your solution is mostly fine, however I would change the line $$(w^{19})^{11*q}.(z^{11})^{19*j}=1^{11*q}1^{19*j}=1$$ to $$(w^{19})^{11*q}.(z^{11})^{19*q}=(-1)^{11*q}(-1)^{19*q}=(-1)^{30*q}=1$$ since we know that $w^{19}$ and $z^{11}$ are $2$nd-primitive roots of unity, and so they are both equal to $-1$ (I'm not sure where the $j$ comes from in your solution).
For the '$\Rightarrow$' part, suppose $(w\cdot z)^{k}=1$. Thus we obtain the equation $$w^{k}=z^{-k}.$$ Raising both sides to the power of $22$, we have that $w^{22k}=(z^{22})^{-k}=1^{-k}=1$, where we have used the fact that $z$ is a $22$nd root of unity. As $w$ is a primitive $38$th root of unity, we see that $22k$ must be divisible by $38$. Since $19$ is a prime which divides $38$ but not $22$, we deduce that $19$ divides $k$.
Similarly, by raising $w^{k}=z^{-k}$ to the power $38$, we see that $38k$ is divisible by $22$, and so $11$ divides $k$ (same reasoning as in the previous paragraph).
Therefore, $k$ is divisible by $11\cdot19=209$, as required.
A: $w=e^{2i\pi m / 38}$ where $\gcd(m,38)=1$.
$z=e^{2i\pi n / 22}$ where $\gcd(n,22)=1$.
Hence, $(wz)^k =  e^{2i\pi k m/ 38}  e^{2i\pi k n/ 22} = e^{2i\pi k (11m+19n)/(2*19*11)} = e^{2i\pi k (11m+19n)/209}$.
Hence, $(wz)^k = 1 \Leftrightarrow k (11m+19n)/209  \in \mathbb{Z} \Leftrightarrow k(11m+19n) \in 209\mathbb{Z} \Leftrightarrow 209 | k(11m+19n) \Leftrightarrow 209 | k$ where the last $\Rightarrow$ follows from $\gcd(m,38)=\gcd(n,22)=1$
A: $w=1^{\frac{1}{38}}$ & $z=1^{\frac{1}{22}}$
$w z=1^{\frac{1}{38}+\frac{1}{22}}\to w z=1^{15/209}$
