List the elements of $\mathbb Z_2 \times \mathbb Z_3$ and write its operation table (the notation is additive). $\mathbb Z_2 \times \mathbb Z_3  = \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}.$
$$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 3) & (1, 1) & (1, 2) & (1, 3) \\
(0, 2)  & (0, 2) & (0, 3) & (0, 4) & (1, 2) & (1, 3) & (1, 4) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (2, 0) & (2, 1) & (2, 3) \\
(1, 1)  & (1, 1) & (1, 2) & (1, 3) & (2, 1) & (2, 2) & (2, 3) \\
(1, 2)  & (1, 2) & (1, 3) & (1, 4) & (2, 2) & (2, 3) & (2, 4) \\
\end{array}
$$
Please, check my work.
 A: I'm terribly sorry to mention it, but I fear this table may be wrong. $\Bbb{Z}_2$ and $\Bbb{Z}_3$ are groups. Therefore, $\Bbb{Z}_2$ X $\Bbb{Z}_3$ is a group.  All such groups are closed, aint they? That is, if a,b is in the set, then a+b (if the operation is additive) is also in the set. Since, for instance, (2,1) is not in the set $\Bbb{Z}_2$ X $\Bbb{Z}_3$ how can it be in the operation table? It seems we would have to restrict the LHS of each pair to modulo 2, so 2 would never appear, would it?
A: $\mathbb Z_3=\{0,1,2\}$
\begin{array}{c|lcr}
+ & 0  &1 & 2\\
\hline
0 & 0  &1 & 2\\
1 & 1  &2 & 0\\
2 & 2  &0 & 1\\
\end{array}
$\mathbb{Z}_3\times\mathbb{Z}_2=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0) \\
(0, 2)  & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (0, 0) & (0, 1) & (0, 2) \\
(1, 1)  & (1, 1) & (1, 2) & (1, 0) & (0, 1) & (0, 2) & (0, 0) \\
(1, 2)  & (1, 2) & (1, 0) & (1, 1) & (0, 2) & (0, 0) & (0, 1) \\
\end{array}
As a general idea, when you write the table group of an operation, each element should appear only once on each column and row. Of course that in $\mathbb{Z}_3, 4=1$ because $4\equiv 1\pmod 3$ but is preferable to keep the same notation for an element.
Edit: The operation on $\mathbb Z_2\times \mathbb Z_3$ is defined as follows: for $(a,b),(c,d)\in \mathbb Z_2\times \mathbb Z_3$, i.e. $a,c\in\mathbb Z_2, b,d\in\mathbb Z_3, (a,b)+(c,d)=(a+c,b+d)$. The idea when you write these elements is to keep the same writing always. 
When you make a sum in $\mathbb Z_n$, the elements are classes of equivalence (in $\mathbb{Z}_n, 0=n=2n=\dots; 1=n+1=2n+1=\dots ; 2=n+2=2n+2=\dots$ and so on - basically whenever an n appears it is made $0$)and you choose to work with the smallest representants $\{0,1,\dots,n-1\}$, so if you have that for example $x+y=n+k$, you shall write $x+y=k$.
A: As Omne Bonum mentions, adding two members of a group must yield a member of the group. Using the notation given in the question, in $\mathbb Z_2 \times \mathbb Z_3$, 
$$(a, b) + (c, d) = ((a+c)\mod 2, \ (b+d)\mod 3)$$
Here is the corrected addition table.
$$\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0) \\
(0, 2) & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (0, 0) & (0, 1) & (0, 2) \\
(1, 1) & (1, 1) & (1, 2) & (1, 0) & (0, 1) & (0, 2) & (0, 0) \\
(1, 2) & (1, 2) & (1, 0) & (1, 1) & (0, 2) & (0, 0) & (0, 1) \\
\end{array}$$
