 Solution

The remaining proper subgroups areTheir orders are respectively one, three, six, four and six. The order-four groups are Klein four-groups; the last two are dihedral groups of order six, isomorphic to $S_3$.

For completeness, if $H$ contains a reflection, write $H\cap ⟨g⟩=⟨g^d⟩$ with $d\mid 6$. Any other reflection differs from a chosen $g^{j}h$ by an element of this rotational subgroup, so $H=⟨g^d,g^{j}h⟩$ with $j$ specified modulo $d$. The choices $d=1,2,3,6$ give respectively the whole group, the two order-six groups, the three order-four groups and the six reflection groups already listed. This proves there are no omitted subgroups.

Thus there are eight remaining proper subgroups. The printed count nine includes one too many; nine would be correct if the whole group were included with these remaining subgroups. Altogether $G$ has sixteen subgroups, fifteen proper.