I think I have some kind of fundamental misunderstanding of direct products and direct sums of groups. I'm trying to understand the following statement:
For prime numbers $p \neq q$, we have $\mathbb{Z} / pq \mathbb{Z} = \langle \bar p \rangle \oplus \langle \bar q \rangle \cong \mathbb{Z} / p \mathbb{Z} \oplus \mathbb{Z} / q \mathbb{Z}$. (This is a version of the Chinese Remainder Theorem.)
For some reason my brain wants the first equality to be $\mathbb{Z} / pq \mathbb{Z} \cong \langle \bar p \rangle \oplus \langle \bar q \rangle$, and the second to be $\langle \bar p \rangle \oplus \langle \bar q \rangle = \mathbb{Z} / p \mathbb{Z} \oplus \mathbb{Z} / q \mathbb{Z}$.
I've tried to construct an example by letting $p = 2$ and $q = 3$, where I believe \begin{align} &\langle \bar 2 \rangle = \{ \bar 0, \bar 2, \bar 4 \} \qquad \text{and} \\ &\langle \bar 3 \rangle = \{ \bar 0, \bar 3 \}, \end{align} so that $\mathbb{Z} / 6 \mathbb{Z} \cong \langle \bar 2 \rangle \oplus \langle \bar 3 \rangle$. I have constructed what I think is the Cayley table of $\mathbb{Z} / 2 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z}$ —
\begin{array}{|c|c|c|c|c|c|} \hline + & (0,0) & (4,3) & (2,0) & (0,3) & (4,0) & (2,3) \\ \hline (0,0) & (0,0) & (4,3) & (2,0) & (0,3) & (4,0) & (2,3) \\ \hline (4,3) & (4,3) & (2,0) & (0,3) & (4,0) & (2,3) & (0,0) \\ \hline (2,0) & (2,0) & (0,3) & (4,0) & (2,3) & (0,0) & (4,3) \\ \hline (0,3) & (0,3) & (4,0) & (2,3) & (0,0) & (4,3) & (2,0) \\ \hline (4,0) & (4,0) & (2,3) & (0,0) & (4,3) & (2,0) & (0,3) \\ \hline (2,3) & (2,3) & (0,0) & (4,3) & (2,0) & (0,3) & (4,0) \\ \hline \end{array}
— as well as what I believe to be the Cayley table of $\mathbb{Z} / 6 \mathbb{Z}$:
\begin{array}{|c|c|c|c|c|c|} \hline + & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 0 \\ \hline 2 & 2 & 3 & 4 & 5 & 0 & 1 \\ \hline 3 & 3 & 4 & 5 & 0 & 1 & 2 \\ \hline 4 & 4 & 5 & 0 & 1 & 2 & 3 \\ \hline 5 & 5 & 0 & 1 & 2 & 3 & 4 \\ \hline \end{array}
If I understand it correctly, a previous theorem in the lectures notes claims that there is an isomorphism $\phi : \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \to \mathbb{Z} / 6 \mathbb{Z}$ defined by $$ \phi(a, b) = a + b , $$ which seems to explain the link between the two Cayley tables.
However, I've tried something similar with $p=3$ and $q=4$, and everything still works out despite $4$ not being prime?? (I realise that primeness is only a sufficient condition.) I feel like I am misunderstanding the difference between direct products and direct sums, or not calculating the groups generated by $\bar 2$ and $\bar 3$ correctly...or I haven't quite understood the $\phi$ isomorphism.
How would you explain the quoted statement above?