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 $\{\mathrm{id}\}$ and the whole group are normal subgroups. The subset $H=\{\mathrm{id},\phi \}$, where $\phi$ 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 $\psi \phi {\psi }^{-1}=\psi \phi \psi$ is the mapping that sends $1$ to $3$, $2$ to $2$, and $3$ to $1$. This bijection does not belong to $H$.