Let $G$ be the group $Z_2$. How would I define the following group, $K$? $K$ consists of ordered pairs of elements of $G$, and the operation on K is $component-wise$ addition. For example: $(a,b) + (c,d) = (a+c, b+d)$. (These sums are taking place mod 2 of course) If I assume, with the operation described above, $K$ really is a group (I can verify this). What would the Cayley table for $K$ look like? And would $K$ be cyclic? I know the group $K$ has 4 elements
 A: $\mathbb{Z \times Z}$ does not refer to the cross product; in abstract algebra, it refers to the cartesian product. Basically, just ordered pairs/triplets/etc. of numbers. So $\mathbb{Z_3 \times Z_5}$ is every ordered pair where the first component is an element of $\mathbb{Z_3}$ and the second component is an element of $\mathbb{Z_5}$. Similarly, $\mathbb{Z_2 \times Z_3 \times Z_5}$ is every ordered pair where the first component is an element of $\mathbb{Z_2}$, the second an element of $\mathbb{Z_3}$, and the third an element of $\mathbb{Z_5}$. $(1, 2, 2)$, for example.
Now, $K = \mathbb{Z_2 \times Z_2} = \{(0,0), (0,1), (1,0), (1,1)\}$
The Caylay table would be:
\begin{bmatrix}
+ && (0,0) & (0,1) & (1,0) & (1,1) \\ \\
(0,0) && (0,0) & (0,1) & (1,0) & (1,1) \\
(0,1) && (0,1) & (0,0) & (1,1) & (1,0) \\
(1,0) && (1,0) & (1,1) & (0,0) & (0,1) \\
(1,1) && (1,1) & (1,0) & (0,1) & (0,0) \\
\end{bmatrix}
If this group were cyclic, the whole group would be additively generated by a single element. We see, though, that this can't be the case:
$$(1,1) + (1,1) = (0,0) \\ (0,0) + (1,1) = (1,1)$$
So $(1,1)$ can't be the group's generator; additive powers of $(1,1)$ cycle between $(1,1)$ and $(0,0)$. By a similar argument, additive powers of $(1,0)$ and $(0,1)$ will just cycle between $(1,0)$ & $(0,0)$ and $(0,1)$ & $(0,0)$ respectively.
