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 $G$?

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    $\begingroup$ The elements of $\Bbb Z_3 \rtimes_{\alpha} \Bbb Z_2$ are made of up pairs $(a^k,b^m)$, so your rows and columns have the wrong headings, and you don't have enough of them (there are $36$ possible products of the $6$ elements). $\endgroup$ – David Wheeler Mar 9 '16 at 3:33
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    $\begingroup$ And as a consistency check when you've redone things, it should bear a striking resemblance to everyone's favorite nonabelian group of order $6$. $\endgroup$ – pjs36 Mar 9 '16 at 3:44
  • $\begingroup$ @DavidWheeler for example $(\bar{2},\bar{2}) + (\bar{0},\bar{1})=(\bar{2},\bar{0})$? $\endgroup$ – user3772119 Mar 9 '16 at 5:17
  • $\begingroup$ You would never write "2 mod 2". So your "sum" should be (2,0) + (0,1) = (2,1). The first coordinate is mod 3. We use the fact that $\Bbb Z_2$ acts automorphically (is that even a word?) on $\Bbb Z_3$ (that is what $\alpha$ is for) to have the second coordinate of the first term "give the first coordinate of the second term a whirl" before we add the two first coordinates. I really, really don't like using $+$ for the semi-direct operation, because it usually is NOT an abelian group. $\endgroup$ – David Wheeler Mar 9 '16 at 11:52

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.

  • $\begingroup$ So for example $(\bar{2},\bar{2}) + (\bar{0},\bar{1})=(\bar{2},\bar{0})$? $\endgroup$ – user3772119 Mar 9 '16 at 4:32
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    $\begingroup$ Not quite: $(\bar{2}, \bar{2}) = (\bar{2}, \bar{0})$ because the second coordinate is mod 2, and then $\alpha_{\bar{0}}$ is the identity map, so $(\bar{2}, \bar{2}) + (\bar{0}, \bar{1}) = (\bar{2}, \bar{1})$. $\endgroup$ – Unit Mar 9 '16 at 14:32

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.

  • $\begingroup$ So for example $(\bar{1},\bar{1}) + (\bar{2},\bar{1})=(\bar{-1},\bar{2})=(\bar{2},\bar{2})$? $\endgroup$ – user3772119 Mar 9 '16 at 4:27
  • $\begingroup$ The second "coordinate" is mod 2, so it should be (2,0), but otherwise, yes. $\endgroup$ – David Wheeler Mar 9 '16 at 11:42

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