Let $G$ a group. Let $x\in G$. Assume that for every $y\in G, xyx=y^3$. Prove that $x^2=e$ and $y^8=e$ for all $y\in G$. To put this in context, this is my first week abstract algebra.
Let $G$ a group. Let $x\in G$. Assume that for every $y\in G, xyx=y^3$.
Prove that $x^2=e$ and $y^8=e$ for all $y\in G$.
A hint would be appreciated.

$\textbf{Full answer:}$ 
Let $x\in G$. Suppose $(\forall y\in G)(xyx=y^3)$.
(i) Set $y=e$ to get $xyx=xex=x^2=e=e^3=y^3$. (The OP already knew this part).

(ii) The hypothesis is $(\forall y\in G)(xyx=y^3)$. Let $y\in G$ be taken arbitrarily. By the hypothesis we have $xyx=y^3$. Cubing both sides we get $(xyx)(xyx)(xyx)=y^9$, i.e., $xy(x^2)y(x^2)yx=y^9$. Therefore $xyyyx=y^9$, that is, $xy^3x=y^9$. Recalling that $y^3=xyx$ it follows $y^9=xy^3x=x(xyx)x=x^2yx^2=eye=y$. From $y^9=y$ you can conclude that $y^8=e$ by multiplying by $y^{-1}$.

