$\mathbb{Z}_{2} \times \mathbb{Z}_{3} = \mathbb{Z}_{6}$

The cyclic group $\mathbb{Z}_{6}$ is the direct product of $\mathbb{Z}_{2}$ and $\mathbb{Z}_{3}$. How are their representations related?

Representations of $\mathbb{Z}_{2}$ and $\mathbb{Z}_{3}$ are related to representations of $\mathbb{Z}_{6}$ by
\begin{equation}
\begin{split}
(0,0) &\to 0, \\
(1,1) &\to 1, \\
(0,2) &\to 2, \\
(1,0) &\to 3, \\
(0,1) &\to 4, \\
(1,2) &\to 5,
\end{split}
\end{equation}
or in general
\begin{equation}
(x,y) \to (a x + b y) \mod 6,
\end{equation}
where $a = 3$ and $b = 4$.

As explained in Concrete Mathematics by Graham, Knuth and Patashnik, the coefficients $a$ and $b$ arise from noting that if $(1, 0) \to a$ and $(0,1) \to b$, then $(x,y) \to (a x + b y) \mod 6$. Thus $a$ and $b$ are found from the equations
\begin{equation}
a = 1 \mod 2, \quad a = 0 \mod 3, \quad b = 0 \mod 2, \quad b = 1 \mod 3.
\end{equation}

We can visualise the relation $\mathbb{Z}_{2} \times \mathbb{Z}_{3} = \mathbb{Z}_{6}$ as
× =

Note that $4 = -2 \mod 6$ and $5 = -1 \mod 6$ is consistent with the representation $4$ of $\mathbb{Z}_{6}$ being the complex conjugate of $2$ and $5$ being the conjugate of $1$.