Pt1420 Unit 6 Assignment 1

The free group with two generators a and b is composed of all finite strings that are formed from the four symbols a, a−1, b and b−1 given that no a is directly next to an a−1 and no b is directly next to a b−1. Two such strings can be concatenated and converted into a string of this type by repeatedly replacing the "forbidden" substrings with the empty string. For example: abab−1a−1 concatenated with abab−1a yields abab−1a−1abab−1a, which contains the substring a−1a underlined in the example, and so the “forbidden” string gets cancelled and reduced to abab−1bab−1a, which again contains the substring b−1b underlined in the example, which then gets reduced to abaab−1a. It can be checked that the group formed from the set of those strings with
For instance, one of the strings in the set aS(a^(-1) ) may be 〗aa^(-1) b which, because of the “forbidden” rule that I discussed earlier which states that a must not appear next to a^(-1), aa^(-1)gets cancelled and reduced to the string b. Similarly, the set contains all the strings that start with 〗a^(-1) (for example the string 〗〖aa〗^(-1) a^(-1) which reduces to〗〖 a〗^(-1)). In this way, the set aS(a^(-1) ) contains all the strings that start with b, b^(-1) and〖 a〗^(-1).
This decomposed the group F2 into four pieces (in addition to the singleton {e} –singleton means a single element in a set-), then by multiplying with a or b two of them were "shifted", then "reassembled" two pieces to make one copy of F_2 and the other two to make another copy of F_2. That is exactly what Banach and Tarski wanted to do to that sphere.
Step 2
In order to find a free group of rotations of 3D space, i.e. that behaves just like (or "is isomorphic to") the free group F2, two orthogonal axes are taken, in this example I will take the x and z axes, and let A be a rotation of θ=arcoss(1/3) about the x axis (arcoss is defined as the one angle between 0 and π radians whose cos in this case is 1/3.), and B be a rotation