$A=\{z:z^{18}=1\}$ and $B=\{z:z^{48}=1\}$, find the number of distinct elements on $C=\{x\times w:x\in A\ \text{and}\ w \in B\}$

Question

If $A=\{z\in \mathbb{C}: z^{18}=1\}$ and $B=\{z\in \mathbb{C}: z^{48}=1\}$ find the number of distinct elements on $$C=\{x\times w: x\in A\ \text{and}\ \ w\in B\}.$$

Source: list of problems for the preparation for math contests

Hints and answers are welcomed. Sorry if this is a duplicate.

Answer 1

Hint: If $c \in C$, then $c^{\operatorname{lcm}(18,48)}=1$.

Spoiler:

$A = \{ \alpha^n : n \in \mathbb Z \}$, where $\alpha$ is a primitive $18$-root of unity. $B = \{ \beta^m : m \in \mathbb Z \}$, where $\beta$ is a primitive $48$-root of unity. Note that $\alpha=\gamma^8$ and $\beta=\gamma^3$, where $\gamma$ is a primitive $144$-root of unity.

Answer 2

To answer this question, it is important to note that if $\eta$ is a $144$th primitive root of unity, then $A$ consists of powers of $\eta^8$ and $B$ of powers of $\eta^3$, since $48 \times 3 = 18 \times 8 = 144$.

Therefore, if $z \in \{ab : a \in A , b \in B\}$ , we have $z = \eta^{8c+3d}$ for some $c,d$ integers. However, note that $8$ and $3$ are co prime, so the set $\{8c+3d : c,d \in \mathbb Z\} = \mathbb Z$ (by Bezout's lemma, it contains $1$, and it is clear that this set is closed under addition and taking the negative sign).

Therefore, $\{ab : a \in A , b \in B\} = \{\eta^l : l \in \mathbb Z\}$. How many elements does that have?