This is probably completely obvious and well known, but I just realized <a, b, c | abcba=1> is a one relation monoid presentation of the free *group* on 2 generators. The relator begins and ends with ‘a’, so bcba=abcb is a two-sided inverse for ‘a’. This allows conjugating the relator, which gives bcbaa=1 and aabcb=1. Thus ‘b’ is invertible too, and repeating the same trick also gives an inverse for ‘c’. Every generator has an inverse, so all cyclic conjugates of the relator equal the identity, so also baabc=1. So the monoid is isomorphic to the group <a, b, c | baabc=1> and its possible to remove the generator ‘c’ since it appears in the relator only once. We’re left with a group with two generators and no relations, so the free group. The monoid generators ‘a’ and ‘b’ map directly to the group generators, but their inverses are accessed in this weird way through ‘c’, where a^-1 = bcba and b^-1 = aabc.