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Prove that if t = xy, then t*x = x*t^(-1)

Here is what I got so far.

Proof:

Since |x| = 2 => x^2 = 1; |y| = 2 => y^2 = 1, then (x^2)(y^2) = 1 => (xy)^2 = 1

Suppose t = xy, then t^2 = (xy)^2 = 1

WTS (want to show) t*x = x*t^(-1)

This group looks like Dihedral group D2n, the operation t*x = x*t^(-1) similar to r*s = s*r^(-1),

we know r^n = s^2 = 1 in D2n, since x^2 = 1, we can let x = s, so x^2 = s^2 = 1. Hence we've shown x = s,

Now I need to show t^n = (xy)^n = 1 so that I can let t = xy = r,

and if we show t = xy = r, then we get exactly the same operation in D2n such that r*s = s*r^(-1) where r = t = xy and s = x

Here is what I stuck, how do I show t^n=(xy)^n=1 ? I only know t^2 = (xy)^2 = 1 from the given condition. Now I need to derive from there and to show that t^n = (xy)^n

Any suggestions?

BTW, I think that I post in the wrong session. I'll make a note next time since I can't delete the post.