I wish to show $D_6 \cdot \text{stab}_{S6}(2) \leq S_6$, where $D_6$ is the dihedral group of order $12$ and $S_6$ is the symmetric group of order $6$.
- I have been able to provide the start of a rough, geometric argument as to why $D_6$ should be considered isomorphic to a subgroup of $S_6$: in labelling the vertices of a hexagon by $1, 2, ..., 6$, one can create a $1-1$ correspondence between the $12$ elements of $D_6$ and a collection $\mathcal{S}$ of $12$ distinct permutations in $S_6$ (see here). In applying Cayley's Theorem, $D_6$ is guaranteed to be isomorphic to a group of permutations. In having created this $1-1$ correspondence, I suspect it logically (?) follows that $\mathcal{S} \subseteq S_6$ is the desired group of permutations, to which $D_6$ is isomorphic. Hence, we have $ D_6 \approx \mathcal{S} \leq S_6.$ (Is there any flaw in this argument)?
- Now, we have that $D_6 \leq S_6$ and $\text{stab}_{S6}(2) \leq S_6.$ To show $D_6 \cdot \text{stab}_{S6}(2) \leq S_6$, I believe I must show that $D_6 \cdot \text{stab}_{S6}(2) = \text{stab}_{S6}(2) \cdot D_6$. However, I am stuck at this step–– how would this proof proceed? Equivalently, I have thought of using the finite subgroup test, but again do not know how to prove the property of closure, i.e., if $\alpha, \beta \in D_6 \cdot \text{stab}_{S6}(2)$, so that $\alpha = \delta_1 \kappa_1$, $\beta = \delta_2 \kappa_2$, where $\delta_1, \delta_2 \in D_6$ and $\kappa_1, \kappa_2 \in \text{stab}_{S6}(2)$, then $ab = \delta_1 \kappa_1 \delta_2 \kappa_2 \in D_6 \cdot \text{stab}_{S6}(2).$