 It can also be obtained from the tree representation of the free monoid by identifying vertices. Consider the case \(n = 2\) of two letters. Since the identifications do not change word length, all identifications are of vertices at the same depth from the root. We perform all identifications simultaneously as follows. At every vertex \(v\), identify \(v_{01}\) and \(v_{10}\) and their subtrees. Whereas before there were \(2^n\) vertices at depth \(n\), now there are \(n+1\). Furthermore instead of a tree we have the upper right quadrant of the plane, that is, \(N^2\), rotated 135 degrees clockwise, with every vertex \(v\) at the top of a diamond whose other vertices are \(v_0\) and \(v_1\) at the next level down, and the identified pair \(v_{01} = v_{10}\) below both.