User:TimothyRias/Group Temp

Example

id (keep it as is) mr (mirror across the bisector of the top vertex) r1 (rotate right) mg (mirror across the bisector of the right vertex) r2 (rotate left) mb (mirror across the bisector of the left vertex)

group table (Column then row) • id r1 r2 mr mg mb id id r1 r2 mr mg mb r1 r1 r2 id mb mr mg r2 r2 id r1 mg mb mr mr mr mg mb id r1 r2 mg mg mb mr r2 id r1 mb mb mr mg r1 r2 id

To clarify the group axioms we consider the group of symmetries of ane equilateral triangle. The elements of the group will be operations that keep the shape of the triangle unchanged. (In the images, the vertices are colored only to visualize the operations). We have:

• Two rotations r₁ and r₂ (rotating the triangle right and left by 120°).
• Reflections (m_r, m_g, and m_b) across the bisector of each vertex. (Labeled by the color of the vertex in the original configuration.)
• Finally, the identity operation id leaving everything unchanged is also a symmetry.

In this example group, the axioms can be understood as follows:

1. The closure axiom demands that any two symmetries can be composed. This is indeed the case – for any two symmetries a and b, we can first perform a and then b and the result will still be a symmetry, written symbolically

   b • a ("perform the symmetry b after performing the symmetry a")

   For example, rotating by 120° left (r₂) and then mirroring across the bisector of the red vertex (m_r) equals mirroring across the bisector of the blue vertex (m_b). Using the above symbols, we have:

   m_g • r₂ = m_r (highlighted in blue in the group table).

2. The associativity constraint is the natural axiom to impose in order to make composing more than two symmetries well-behaved: given three elements a, b and c of G, there are two possible ways of computing "a after b after c". The requirement:

   (a • b) • c = a • (b • c)

   means that composing a after b, and calling this symmetry x, then x after c is the same as a after y, where y in turn is applying b after c. For example, we check (m_r • m_b) • r₂ = m_r • (m_b • r₂) using the group table at the right:

   (m_r • m_b) • r₂ = r₂ • r₂ = r₁, which equals

   m_r • (m_b • r₂) = m_r • m_g = r₁.

3. The identity element is the symmetry id leaving everything unchanged: for any symmetry a, performing id after a (or a after id) equals a, in symbolic form:

   id • a = a, and

   a • id = a.

4. Inverse elements are fulfilling the purpose of undoing the operation of some element. In the symmetry group example, every symmetry can be undone: the identity id, the reflections m_r, m_g, and m_b are their own inverse, because repeating them brings the square back to its original orientation. The rotations r₁ and r₂ are each other's inverse, because rotating one way and then the other way leaves the square unchanged. In symbols for example:

   m_b • m_b = id,

   r₂ • r₁ = r₁ • r₂ = id.

Given their existence, both identity element and inverse elements are unique, see the notations section below.