# 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.

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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