Take a permutation g in a group G:

Decompose g as a product of generators of G (negative integers represent inverse generators):

Reconstruct the permutation from the word:

Decompose a permutation as a product of the default generators of a symmetric group:

Decompose the same permutation using a different set of generators for the same group:

The returned words are usually not optimal:

This is the optimal solution:

Check them:

Check that the mapping relating respective generators of these two groups is a homomorphism:

This implements the homomorphism, extending the mapping of generators to all elements:

Check that for any two elements of G1, the homomorphism property is obeyed:

In this case, the homomorphism is actually an isomorphism:

GroupElementFromWord reconstructs the original group element from the word:

Identity generators are not used, but affect the indexing of the other generators:

The identity permutation always corresponds to the empty word:

The element must belong to the group:

The group of a 3×3×3 Rubik's cube is usually given in terms of six generators:

However, one of them is redundant, and the group can actually be constructed with only five face rotations:

This is a possible word expressing the sixth rotation in terms of the first five and their inverses:

Check it:

After removing the generators corresponding to opposite faces, the remaining four still move all facelets, but now they only generate a subgroup of index 2048:

This is because there are two disconnected sets of edge facelets:

Compare with the original situation: