  It is known that if every group satisfying an identity of the form yx ~
xU(x,y)y is abelian, so is every semigroup that satisfies that identity.
Because a group has an identity element and the cancellation property, it is
easier to show that a group is abelian than that a semigroup is. If we know
that it is, then there must be a sequence of substitutions using xU(x,y)y ~ yx
that transforms xy to yx. We examine such sequences and propose finding them as
a challenge to proof by computer.
  Also, every model of y ~ xU(x,y)x is a group. This raises a similar
challenge, which we explore in the special case y ~ x^my^px^n. In addition we
determine the free model with two generators of some of these identities. In
particular, we find that the free model for y ~ x^2yx^2 has order 32 and is the
product of D4 (the symmetries of a square), C2, and C2, and point out relations
between such identities and Burnside's Problem concerning models of x^n= e.
