How to compute in $\Bbb{Z}_6 \rtimes \Bbb{Z}_2$?

Question

I had an exercise in the algebra lecture, where I had to show that

$$\Bbb{Z}_6 \rtimes \Bbb{Z}_2 \approx S_3 \times \Bbb{Z}_2.$$

One way to solve this exercise is to find an element of order $6$ in $S_3 \times \Bbb{Z}_2$, for instance $((132),1)$. This element generates a subgroup of $S_3\times \Bbb{Z}_2$, which has index $2$, hence it is normal in $S_3 \times \Bbb{Z}_2$. Now we can find an element in $S_3 \times \Bbb{Z}_2$ of order $2$, for instance $((12),0)$. If we let $H := \langle ((132),1) \rangle$, and $K := \langle ((12),0)\rangle$ then $K\lhd H$. Moreover one can see that $H\cap K = \{(id,0)\}$ hence $H \rtimes S \approx S_3 \times \Bbb{Z}_2$.

I am wondering if it is possible to show that the two groups are isomorphic by computing their multiplicative table, since they are "small" one can think to evaluate the multiplicative table of $\Bbb{Z}_6 \rtimes \Bbb{Z}_2$ and the one of $S_3 \times \Bbb{Z}_2$ and find an isomorphism directly.

It is no problem to evaluate the multiplicative table of $S_3\times \Bbb{Z}_2$ because the multiplication there is just the component-wise multiplication.

I don't know how to compute in $\Bbb{Z}_6 \rtimes \Bbb{Z}_2$. I know that the multiplication there is defined like this: let $(a_1,b_1),(a_2,b_2) \in \Bbb{Z}_6 \rtimes \Bbb{Z}_2$ then $$(a_1,b_1)(a_2,b_2):= (a_1 \varphi_{b_1}(a_2),b_1b_2)\qquad \qquad (1)$$

where $\varphi_{b_1} \in \text{Aut}(\Bbb{Z}_6)$.

I wonder how is the isomoriphism $\varphi_{b_1}$ defined. I could find $4$ isomorphism from $\Bbb{Z}_6 \to \Bbb{Z}_6$ (I think there are not more, but i might be wrong). The question is now, how do i define such an isomorphism? Which one should i take in order to evaluate $(1)$?

Answer 1

You seem to be confusing matters. A semidirect product $G\rtimes_\eta H$ needs a group morphism $\eta:H\to {\rm Aut}G$. In your case you want the semidirect product to be nonabelian, and since ${\rm Aut}\; \mathbb Z_6=\mathbb Z_2$ you need to be taking the only nontrivial morphism $\eta:\mathbb Z_2\to\mathbb Z_2$ that sends the generator of $\mathbb Z_2$ to the nontrivial automorphism of $\mathbb Z_6$, inversion. This sends $g\to g^{-1}$.

Thus in $\mathbb Z_6\rtimes \mathbb Z_2$, let $h$ denote the generator of $\mathbb Z_6$ and $g$ that of $\mathbb Z_2$. We have $ghg^{-1}=\eta(g)(h)=h^{-1}$. This should allow you to compute every other product, since this is generated by $g$ and $h$.

Answer 2

I would first try to identify the element $(id_{S_3}, 1)$ in $\mathbb{Z}_6 \rtimes \mathbb{Z}/2$. Since the former element is central, we seek an element in the latter which commutes with every element.

Since $(n,1)$ does not commute with any element except itself (pretty easy to check), and $(n, 0)$ only commutes with $(0, 1)$ if $n = 3$, it follows that our central element is the element $(3, 0) \in \mathbb{Z}/6 \rtimes \mathbb{Z}/2$.

So you now know that $((123), 1) \mapsto (1, 0)$ and that $(id_{S_3}, 1) \mapsto (3, 0)$.

The only thing left to do is to determine where a transposition goes. Since this must have as image an element of order two. There are quite a few of these (for example, $(n, 1)$ is always of order 2). You should be able to play around and find an isomorphism that works (it will not be unique!)