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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.

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up vote 2 down vote accepted

Hint: The object $x$ is fixed, I would rather call it $a$. I assume you have proved that $a^2=e$.

For any object $u$, we have $aua=(u)(u)(u)$. Let $u=aya$.

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I don't understand how you defined $y$ here. – Kasper Feb 18 '13 at 20:16
I get $y=xy^3x$. – Kasper Feb 18 '13 at 20:20
But $xy^3x$ is $y^9$. I would do the same thing a bit differently but equvalently, getting $y=(xyx)(xyx)(xyx)=(y^3)(y^3)(y^3)$. – André Nicolas Feb 18 '13 at 20:41

HINT The relation holds for every element in the group. What special element does every group have?

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ah, of course, $xex=eee=e \implies x^2=e$ – Kasper Feb 18 '13 at 20:04

$\textbf{Full answer:}$

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 $y^{-1}$.

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