= Solution
The proper <normal subgroups> and their <quotient groups> are as follows. Here the trivial subgroup is included as a proper subgroup.
* $N=\{1\}$ gives $G/N\cong D_{12}$.
* $N=\langle g^3\rangle$ gives $G/N\cong S_3$: the rotation has order three and the reflection inverts it.
* $N=\langle g^2\rangle$ gives $G/N\cong C_2\times C_2$: the two generators now have order two and commute.
* $N=\langle g\rangle$ gives $G/N\cong C_2$.
* $N=\langle g^2,h\rangle$ or $\langle g^2,gh\rangle$ gives $G/N\cong C_2$.
The rotational subgroups are preserved by reflection conjugation, and the last three order-six subgroups are normal by index two. The three order-four groups are not normal, since conjugation by $g$ changes the reflection exponent by two, which is nonzero modulo three. Part <i/solution> excluded the reflection subgroups. These observations, together with the complete subgroup list, prove the normal-subgroup list is exhaustive.
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