I am trying to visualizing $S_3 \rtimes D_4$ following this video. Here, $S_3$ is the symmetric group over three symbols and $D_4$ is the dihedral group of order $8$. The semidirect product is defined as follows as per the definition given here.
Outer semidirect product:
Let $G$ be $G = S_3 \rtimes_\varphi D_4$, $S_3 \triangleleft G$, and $D_4 < G$. Let $Aut(S_3)$ denote the group of all automorphism of $S_3$. The map $\varphi : D_4 \to Aut(S_3)$ defined by $\varphi (h) = \varphi_h$, conjugation by $h$, where $\varphi (h) (n) = \varphi_h (n) = h n h^{-1}$ for all $h$ in $D_4$ and all $n$ in $S_3$, is a group homomorphism.
Together $S_3$, $D_4$, and $\varphi$ determine $G$ up to isomorphism.
So, $G = S_3 \rtimes_\varphi D_4$ is defined as follows.
- The underlying set is the Cartesian product $S_3 \times D_4$
- The operation, $\cdot$, is determined by the homomorphism, $\varphi$: $$ \cdot : (S_3 \rtimes_\varphi D_4) \times (S_3 \rtimes_\varphi D_4) \to (S_3 \rtimes_\varphi D_4) \text{ s.t. }\\ (n_1, h_1) \cdot (n_2, h_2) = (n_1, \varphi(h_1)(n_2), h_1h_2) = (n_1, \varphi_{h_1}(n_2), h_1h_2) $$ for $n_{1,2} \in S_3$ and $h_{1,2} \in D_4$.
Now, I go by each concept as it is done in the video.
To do that we need to know the Cayley graph of $S_3$ and $D_4$.
$S_3$:
According to this wiki article $Aut(S_n) = S_n$. So, the Cayley graph of $Aut(S_3)$ will be same as above.
$D_4$:
This is where I am stuck. How can one guess the homomorphism $\varphi : D_4 \to Aut(S_3)$ just by eyeballing the these two Cayley graphs?