# Presentation of symmetric group

In a textbook, I read the following:

We have already run into several examples of presentations. For instance, the free group $F(A)$ is presented by $(A\mid\emptyset)$ [ . . . ] $S_3$ as a quotient of the free group $F(\{x,y\})$ by the smallest normal subgroup containing $x^2, y^3$ and $yx = xy^2$: $(x,y\mid x^2,y^3,xyxy)$ in shorthand.

I am unsure why the condition $yx = xy^2$ is represented as $xyxy$ in the presentation? I need to transform it to an equation of the form $xyxy = 1$. But I don't succed, for example if I use the facts $x = x^{-1}$ and $y^{-1} = y^2$ which follow from $x^2 = 1$ and $y^3 = 1$ and substituting I got $yx = x^{-1}y^{-1}$. Now multiplying by $y$ and $x$ from the right yields $yxyx = 1$, but this is in general different from $xyxy = 1$, so why should it be the same?

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$xyxy=1$ is equivalent to $yxyx=1$. Multiply both sides of $xyxy=1$ on the left by $x^{-1}$, then multiply both sides on the right by $x$. –  Ted Feb 18 '13 at 2:18

You are on the right track, you just have to multiply by $x$ on the left and by $y$ on the right.
You correctly took into account that $x=x^{-1}$ and $y^2=y^{-1}$. Hence $yx=xy^2$ if and only if $xyx=y^{-1}$ if and only if $xyxy=1$.