Let $G = S_3$ and $a=(12)$, $b=(13)$ and $c=(23)$. Let $H=\langle a\rangle$ if $(\cdot)$ is the group action of conjugation of $G$ in $H$ then find $a\cdot H$, $b\cdot H$ and $c\cdot H$
My attempt
Let $\cdot : G\times H \rightarrow H$ sucht that $(g,h)\rightarrow g*h*g^{-1}$ the action of conjugation, then for $g=a=(12)$ we have:
$$a*h*a^{-1}=(12)\circ(12)\circ(21)=(12)$$
For $b\cdot H$ we have:
$$b*h*b^{-1}=(13)\circ(12)\circ(31)=(32)$$
For $c\cdot H$ we have:
$$c*h*c^{-1}=(23)\circ(12)\circ(32)=(13)$$
Then $a\cdot H= \{(12),(1)(2)(3)\}$, $b\cdot H= \{(32),(1)(2)(3)\}$, $c\cdot H= \{(13),(1)(2)(3)\}$
Is this correct?