Is t*x = x*t^(-1) for Elements of Order 2 in Any Group?

[xsw001]

Let x, y be elements of order 2 in any group G.
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.

 

[Delta2]

Notice that $$t^{-1}=y*x$$ because $$t*(y*x)=(t*y)*x=(x*(y*y))*x=(x*1)*x=1$$and similarly $$(y*x)*t=1$$.

 

[xsw001]

Yes! Got it!
Since t^-1=(y^-1)*(x^-1) and x^2=1 => x=x^-1, y^2=1 => y=y^-1, so t^-1=yx
Then tx=xyx, and x(t^-1)=xyx, indeed they are equal!
Thanks for the tips Delta! Now it seems so simple!

 

[Delta2]

Glad to help !

 

[blue_raver22]

No problem, let's continue with the proof.

To show that t^n = (xy)^n = 1, we can use the fact that t^2 = (xy)^2 = 1. This means that t^2 is the identity element in the group G.

Now, let's consider t^3 = (xy)^3 = (xy)(xy)(xy). Using associativity, this can be rewritten as x(yx)(yx)y. Since x has order 2, we know that x*x = 1. Similarly, y*y = 1. So, we can rewrite t^3 as x(1)(1)y = xy.

But we know that t^2 = 1, so we can substitute this in and get t^3 = t*t^2 = t*1 = t. Therefore, t^3 = t = xy.

Similarly, we can show that t^4 = (xy)^4 = 1. This follows from the fact that t^2 = 1, so t^4 = t^2*t^2 = 1*1 = 1.

In general, we can show that t^n = (xy)^n = 1 for any positive integer n. This is because we can keep using the fact that t^2 = 1 and t^n = t*t^(n-1) to rewrite t^n as t^(n-2)*t^2 = t^(n-2)*1 = t^(n-2) = ... = 1.

So, we have shown that t^n = (xy)^n = 1 for all positive integers n. Now, we can let t = xy = r, and use the same reasoning as in Dihedral group D2n to show that t*x = x*t^(-1) = x*r^(-1) = x*(xy)^(-1) = x*y^(-1)*x^(-1) = x*y*x*x = x*(1)*x = x*x = 1.

Therefore, we have proven that t*x = x*t^(-1) for any elements x, y of order 2 in any group G.