Showing elements are self-inverse

Question

Hi,

I need help show algebraically that this is true. If G is a finite group and an element x has order 3 and element y has order 2. Show that if

yx=x²y then elements xy and x²y are both self-inverse.

thank you so much

Ruby 😀

Answer 1

First, we show that yx is a self-inverse (and, thus, x²y is also a self-inverse).

Let e be the identity element. Then, as x is of order 3 and y is of order 2, x³ = e and y² =e

Then, we show that x²y is the inverse of yx

yx = x²y, yxx²y = yx³y = yey = yy =e, and x²yyx = x²ex = x²ex = x²x = x³ = e

Thus yx and x²y are inverses of each other. Yet, it is given that they are equal. Thus, each is a self-inverse.

Then, consider xy

xyxy = exyxy = (yxx²y)xyxy = yxx²(yxyx)y = yxx²y = e, so xy is a self-inverse