Let $S = \{A,B,C.D,E \}$ be the points located on the square in Figure $2.$ Let $D_4 = \{1,r,r^2,r^3, s, rs, r^2 s , r^3 s\ |\ r^4 = 1, s^2 = 1, rs = sr^{-1} \}$ be the dihedral group, where $r$ is the rotation by $90$ degree and $s$ is the reflection along the main diagonal. Then $D_4$ acts on $S$ which also gives an action of $D_4$ on the power set $\mathcal P (S).$ Which of the following sets is fixed by the action of $D_4\ ?$
$(1)$ $\{A,E,D \}$
$(2)$ $\{D,E, B \}$
$(3)$ $\{\{A,C\}, \{E\}, \{D,B\} \}$
$(4)$ $\{\{A,D\}, \{E\} \}$
Clearly $(1)$ and $(2)$ are not fixed points under this action because the reflection about $DB$ takes the set $\{A,E,D\}$ to the set $\{C,E, D\}$ and the reflection about the horizontal axis passing through $E$ will take the set $\{D,E,B\}$ to the set $\{A,E,C\}.$ The last option $(4)$ is also false because the vertical flip will take the set $\{\{A, D\}, \{E\}\}$ to the set $\{\{B,C\}, \{E\} \}.$ So the one which is left i.e. $(3)$ has to be the correct option. Am I right?
Kindly check my argument above. Thanks in advance.