As the title says, I would like to find all the groups $G$ of order $12$ such that there exists a normal subgroup $H \subseteq G$ with $G/H \simeq S_3$ ($S_3$ is the symmetric group of degree $3$).
The following is my line of thought, and I would like you to check if this is correct:
I need to check if the possible choices of $G$ can have such a normal subgroup $H$, and then check if $G/H \simeq S_3$. Due to cardinality reasons, such a subgroup $H$ must be isomorphic to $C_2$ (cyclic group of order 2).
The groups of order $12$ are the following (cf. Groupprops):
- $C_{12}$ (cyclic group of order $12$),
- $D_6$ (dihedral group of order $6 \cdot 2$),
- $A_4$ (alternating subgroup of $S_4$),
- $Dic_{12}$ (dicyclic group of order $12$),
- $C_6 \times C_2$ (direct product of $C_6$ and $C_2$).
Since $S_3$ is not abelian, we can exclude the groups of order $12$ which are abelian (in this case: $C_{12}$ and $C_6 \times C_2$.
According to Groupprops, the only normal subgroups of $A_4$ are $\{ e \}, V_4, A_4$ (where $V_4$ is the Klein four group). None of them have cardinality $2$, so $A_4$ is also not possible.
We can write $D_6 = C_6 \rtimes C_2$, and if $r$ is a generator of $C_6$, we can take $H = \langle r^3 \rangle \times \{ e \} \simeq C_2$. With that choice, we have $G/H \simeq C_3 \rtimes C_2 = S_3$ as desired.
Using Groupprops again, we see that $Dic_{12} = C_3 \rtimes C_4$. If $x$ is a generator of $C_4$, choose $H = \{ e \} \times \langle x^2 \rangle$, it should be $G/H \simeq C_3 \rtimes C_2 = S_3$ again.
Therefore, $D_6$ and $Dic_{12}$ are the only two possible choices.
Could you please confirm if this is correct?