Permutation group S3/Subgroups and normal subgroup/Example

We consider the permutation group ${}G=S_{3}$ for a set with three elements, that is, ${}S_{3}$ consists of all bijective mappings of the set ${}\{1,2,3\}$ to itself. The trivial group ${}\{\operatorname {id} \}$ and the whole group are normal subgroups. The subset ${}H=\{\operatorname {id} \,,\varphi \}$, where ${}\varphi$ is the element that swops ${}1$ and ${}2$ and fixes ${}3$, is a subgroup. However, it is not a normal subgroup. To show this, let ${}\psi$ denote the bijection that fixes ${}1$ and swops ${}2$ and ${}3$. The inverse of ${}\psi$ is ${}\psi$ itself. The conjugation ${}\psi \varphi \psi ^{-1}=\psi \varphi \psi$ is the mapping that sends ${}1$ to ${}3$, ${}2$ to ${}2$, and ${}3$ to ${}1$. This bijection does not belong to ${}H$.