Definition. Given a group $K$ with subgroups $G$ and $H$ satisfying (i) $K = GH$, i.e., every element $k \in K$ can be written $k = gh$ where $g \in G$ and $h\in H$ (ii) $G K$ (iii) $G\vartriangleleft H = {1}$ we then say that $K$ is the semidirect product $G \rtimes H$.
Definition. $\Sigma_l$ acts on $\mathbb{E} = \mathbb{R}^l = \{(X_l,..., X_l)\}$ by permuting coordinates. We thereby obtain a subgroup $\Sigma_l \subset O(\mathbb{E})$. $O(\mathbb{E}) = \{f: \mathbb{E} \to \mathbb{E} linear and (f(x),f(y))= (x,y) for all x,y \in \mathbb{E} \}$ be the orthogonal group of lEThis subgroup is a reflection group.
I found that $\mathbb{Z}/ 2\mathbb{Z}$ is generated by the involution $(1,2)$. But I don't understand why it is involution? And how I can write this formula? Also I found that $\mathbb{Z}/ 3\mathbb{Z} $ is generated by the cycle $(1,2,3)$. But why we have $\Sigma_3 =\mathbb{Z}/ 3\mathbb{Z} \rtimes \mathbb{Z}/ 2\mathbb{Z}$?
Cycle $(1,2,3)$ means that $1 \to 2$, $2\to 3$ and $3 \to 1$.
The underlying set of a semidirect product is the direct product of the underlying sets, so $\mathbb{Z}/ 3\mathbb{Z} \rtimes \mathbb{Z}/ 2\mathbb{Z}$ has order 6.
By definition of multiplication in a semidirect product, $(a, b) * (a' , b') = (a \phi(b)(a') , b b')$, where $\phi : \mathbb{Z}/ 2\mathbb{Z} \rightarrow Aut(\mathbb{Z}/ 3\mathbb{Z})$ send $0$ to the identity and $1$ to negation.
We have $(1, 1) * (2, 1) =(1 +\phi(1)(2), 0) = (1-2, 0) = (2, 0)$
while $(2, 1)*(1, 1) = (2 + \phi(1)(1), 0) = (2-1, 0) = (1, 0)$
Therefore the group is not abelian. The only non-abelian group of order 6 is $S_3$.