Given $x,y\in G$ and $x\neq e$, $|y|=2$ and $yxy^{-1}=x^2$ then find $|x|$

Question

Question: Let $x$, $y$ belong to a group $G$. Assume $x\neq e$, $|y|=2$ and $yxy^{-1}=x^2$ then find $|x|$

My attempt:

I know that $|yxy^{-1}|=|x|$ so by given condition I have $|x|=|x^2|$. Further $(yxy^{-1})^k=yx^ky^{-1}$. Is $|x|$ infinite?

Please Help.

Answer 1

$y^2xy^{-2}=y(yxy^{-1})y^{-1}=yx^2y^{-1}=(yxy^{-1})^2=x^4$

Also, $y^2xy^{-2}=exe=x$ .

Therefore $x^4=x$ and $x^3=e$. Since $x\ne e$ the order of $x$ is $3$.

Answer 2

No, the order of $x$ need not be infinite. Take the dihedral group $$ D_3=\langle x,y\mid x^3=e,y^2=e,xy=yx^2\rangle. $$ We have $y=y^{-1}$ so that $yxy^{-1}=x^2$ and $y$ has order $2$, whereas $x$ has order $3$.