Cayley table for semidirect product $\mathbb{Z}_3 \rtimes _\alpha \mathbb{Z}_2$? 
Let $\alpha : \mathbb{Z}_2 \rightarrow Aut(\mathbb{Z}_3) \cong
\mathbb{Z}_2$ be the homomorphism given by $\alpha_{\bar{1}}(\bar{1})
= \bar{2} \in \mathbb{Z}_3$. Write down the Cayley table for $G=\mathbb{Z}_3 \rtimes _\alpha \mathbb{Z}_2$.

So letting $\mathbb{Z}_3 = \{e,a,a^2\}$ and $\mathbb{Z}_2 = \{e,b\}$, I came up with the table:
\begin{array}{|c|c|c|}
\hline
*& e & a & a^2 \\ \hline
e & (e,e) & (e, a) & (e, a^2)\\ \hline
b & (b,e) & (b,a^2) & (b,a)\\ \hline
\end{array}
I'm not sure this is right, however. I don't think I properly used the expanded definition of $\alpha$ to generate this table. What's the proper Cayley table for $G$?
 A: I can't make tables, but I can tell you what to do. You have omitted many elements! $|G| = 2\cdot 3 = 6$ so you should have a six-by-six table. The elements of $G$ are pairs $\{\bar{0}, \bar{1}, \bar{2}\}\times\{\bar{0}, \bar{1}\}$ and the "multiplication" (funny-looking addition, in this case) is given by $$(a, b) + (a', b') = (a+\alpha_b(a'), b+b').$$
Can you work it out from here? 
I also suggest not redefining $\mathbb{Z}_3 = \{e, a, a^2\}$ and $\mathbb{Z}_2 = \{e, b\}$ when you are already given their descriptions, unless you also translate the definition of $\alpha$ into this new notation.
A: You need to use the definition of the operation in a semi-direct product:
$(a^k,b^m)\ast(a^{k'},b^{m'}) = (a^k[\alpha(b^m)](a^{k'}),b^{m+m'})$
It may be helpful to note that:
$\alpha(b^m) = \text{id}_{\Bbb Z_3}$ if $m$ is even, and
$\alpha(b^m) = \text{inv}$, where $\text{inv}(a^k) = a^{-k}$ (the inversion map), if $m$ is odd.
If we write the operation in the cyclic groups additively (as integers modulo $3$ and $2$) this becomes:
$(k,m)\ast(k',m') = (k+(-1)^mk' (\text{mod }3), m+m' (\text{mod }2))$
It may also be helpful to note that elements of the form $(k,0)$ are "different" than elements of the form $(k,1)$. One of them forms a (normal) subgroup.
