Solution to puzzle 129: Abelian group

Let G be a group with the following two properties:

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

Prove that G is abelian.

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A number of different solutions are possible.

Solution 1

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.

Solution 2

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

x2y	= ((xy−1)y)2y
	= (y(xy−1))2y (by property (i))
	= (yxy−1)(yxy−1)y
	= yx2 (iii)

Then we have:

x−1y−1x	= x(x−1)2y−1x
	= xy−1(x−1)2x (by (iii))
	= xy−1x−1 (iv)

Finally we obtain:

(xyx−1y−1)2	= xy(x−1y−1x)yx−1y−1
	= xy(xy−1x−1)yx−1y−1 (by (iv))
	= xyx(y−1x−1y)x−1y−1
	= xyx(yx−1y−1)x−1y−1 (by (iv), with x, y transposed)
	= (xy)2(x−1y−1)2
	= (yx)2(yx)−2 (by (i))
	= e

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

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

Source: To be announced

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