 This view is not particularly insightful. The group counterpart of the tree representation does a better job of presenting a free group. Consider the free group on \(n = 2\) generators \(A\) and \(B\). We start with the free monoid on 4 generators \(A, B, a, b\) where \(a\) is the inverse of \(A\) and \(b\) that of \(B\). Every vertex of this tree has 4 descendants. So the root has degree 4 and the remaining vertices have degree 5: every vertex except the root has one edge going in, say the generator \(a\), and four out. Consider any nonroot vertex \(v\). The effect of deleting adjacent complementary pairs is to identify the immediate ancestor of \(v\) with one of the four descendants of \(v\), namely the one that makes the path from the ancestor to the descendant a complementary pair. For every nonroot vertex \(v\) these identifications reduce the degree of \(v\) from 5 to 4. The root remains at degree 4.