Let G be a group with the following two properties:

- (i) For all x, y in G, (xy)
^{2}= (yx)^{2}, - (ii) G has no element of order 2.

Prove that G is abelian.

A number of different solutions are possible.

Let e be the identity element. Consider any two group elements x, y.

We begin with property (i): (xy)^{2} = (yx)^{2}.

Hence (xy)^{−1}(xy)^{2}(yx)^{−1} = (xy)^{−1}(yx)^{2}(yx)^{−1}.

That is, (xy)(yx)^{−1} = (xy)^{−1}(yx).

Squaring both sides, we obtain ((xy)(yx)^{−1})^{2} = ((xy)^{−1}(yx))^{2}.

Then, by property (i): ((xy)(yx)^{−1})^{2} = ((yx)(xy)^{−1})^{2}.

Since ((yx)(xy)^{−1})^{−1} = (xy)(yx)^{−1}, we deduce that ((xy)(yx)^{−1})^{4} = e.

Writing this as [((xy)(yx)^{−1})^{2}]^{2} = e, by property (ii) we have ((xy)(yx)^{−1})^{2} = e.

Using (ii) once more, we obtain (xy)(yx)^{−1} = e.

Therefore, xy = yx; that is, G is abelian.

Let e be the identity element. For any two group elements x, y, we have:

x^{2}y | = ((xy^{−1})y)^{2}y |

= (y(xy^{−1}))^{2}y (by property (i)) | |

= (yxy^{−1})(yxy^{−1})y | |

= yx^{2} (iii) |

Then we have:

x^{−1}y^{−1}x | = x(x^{−1})^{2}y^{−1}x |

= xy^{−1}(x^{−1})^{2}x (by (iii)) | |

= xy^{−1}x^{−1} (iv) |

Finally we obtain:

(xyx^{−1}y^{−1})^{2} | = xy(x^{−1}y^{−1}x)yx^{−1}y^{−1} |

= xy(xy^{−1}x^{−1})yx^{−1}y^{−1} (by (iv)) | |

= xyx(y^{−1}x^{−1}y)x^{−1}y^{−1} | |

= xyx(yx^{−1}y^{−1})x^{−1}y^{−1} (by (iv), with x, y transposed) | |

= (xy)^{2}(x^{−1}y^{−1})^{2} | |

= (yx)^{2}(yx)^{−2} (by (i)) | |

= e |

Since G has no elements of order 2, we conclude that xyx^{−1}y^{−1} = e.

Therefore, xy = yx; that is, G is abelian.

Source: To be announced