Symmetry group of a triangular lattice

Question

The question is:

Let $G$ be the group of symmetries of an equilateral triangular lattice $L$. Find the index in $G$ of the subgroup $T \cap G$. ($T$ is the group of translations)

The solution says that $G$ is $D_6$ (dihedral group) instead of $D_3$, so the answer is the order of $D_6$, which is $12$. Why isn't $G$ $D_3$?

Answer 1

It's $D_6$ instead of $D_3$ because it is a triangular lattice. You can rotate the entire lattice around any single vertex by $60^\circ$, and each lattice point maps to another lattice point. In the triangular lattice below, rotation around the black lattice point by any multiple of $60^\circ$ will permute the blue lattice points among themselves and similarly the green lattice points. The composition of six such rotations, not three, takes all the points back to their original positions.

Similarly you can reflect the lattice around any of the six lines of symmetry that pass through a single vertex. (In the illustration above, three are pink, three red.)

Answer 2

Do you mean that $G/T \simeq D_6$? In that case the reason you get $D_6$ instead of $D_3$ is because there are (up to translation) $6$ lines that you can reflect about, not $3$. Imagine an equilateral triangle with a horizontal base. Its symmetry group $D_3$ does not contain reflection about any horizontal line, but if this triangle is part of a lattice then the lattice can be reflected about that line.

Answer 3

Because a symmetry can make a specific triangle upside-down, which can not be killed by modulo translantion.