D₃ is the dihedral group on the six symmetry transformations of an equilateral triangle. (Besides the identity symmetry, it involves two rotations and three reflections.):

1. Given a basic position (edge at bottom, vertex at top), a rotation of 120° = 2p/3 radians sends the triangle back into itself; denote this symmetry as R₁.
2. Given the original basic position, a rotation of 240° = 4p/3 radians sends the triangle back into itself; denote this symmetry as R₂.
3. A reflection -- about the median connecting bisection of basic bottom edge with bisection of vertex angle at top -- sends the triangle back into itself; denote this symmetry as M₁.
4. Another reflection -- about the median connecting bisection of basic right edge with bisected left angle -- sends the triangle back into itself; denote this symmetry as M₂.
5. A third reflection -- about the median connecting bisection of basic left edge with bisected right angle -- sends the triangle back into itself; denote this symmetry as M₃.
6. Finally, the identity symmetry, in which it remains in its basic position, without any change; denote this symmetry as I.

These six symmetry transformations are the operands of this group. Its operation is concatenation (which we'll denote by o), that is, one symmetry, S_i, followed by another one, S_j: S_i o S_j.

The group property is revealed by the result that S_i o S_j = S_k. That is, the concatenation of two transformations has the same effect on the basic position as a single one does -- which means closure of the transformation set. Also, it can be shown that each transformation has an inverse, which "undoes" its effect -- that is, the concatenation of a transformation and its inverse is equivalent to the identity transformation.

We can show this its Group Table:

CAYLEY TABLE OF D₃

o	I	R1	R2	M1	M2	M3
I	I	R1	R2	M1	M2	M3
R1	R1	R2	I	M2	M3	M1
R2	R2	I	R1	M3	M1	M2
M1	M1	M3	M2	I	R2	R1
M2	M2	M1	M3	R1	I	R2
M3	M3	M2	M1	R2	R1	I

This group is said to be of order 6, since it has 6 members. Let's look at its subgroups:

• from the Table, it is clear that the identity I is a subgroup all by itself {I}, of order 1;
• from the Table, we see that M₁ is its own inverse -- that is, M₁ o M₁ = I -- hence, {I, M₁} constitutes a subgroup of this group, of order 2, and the same for the other two medians, with {I, M₂} forming a subgroup of this group, also of order 2, and {I, M₃} forming a subgroup of this group, also of order 2;
• from the Table, we see that R₁, R₂ are mutually inversive -- that is, R₁ o R₂ = R₂ o R₁ = I -- so {I, R₁, R₂} is a subgroup of this group, of order 3;
• however, we see, from the Table, that putting any median into a 2-group involving a different median does not result in closure, which means that there is only the one subgroup of order 3;
• and putting a median in with the rotation subgroup results in closure being found with all the elements of the group, so there is no subgroup of order 4, none of order 5;
• trivially, the group is a subgroup of itself, of order 6;
• and this is all.

So we have subgroups of order 1, 2, 3, 6. This agrees with a well known result in group theory:

Theorem (Lagrange): Given a group of order n, the orders of its subgroups are factors of n.

In Summary, we have subgroups: {I}, {I, M₁}, {I, M₂}, {I, M₃}, {I, R₁, R₂}, {I, R₁, R₂, M₁, M₂, M₃}.