<s>[INST] Deriving a relation in a group based on a presentation Suppose I have the group presentation $G=\langle x,y\ |\ x^3=y^5=(yx)^2\rangle$. Now, $G$ is isomorphic to $SL(2,5)$ (see my proof here).  This means the relation $x^6=1$ should hold in $G$.  I was wondering if anyone knows how to derive that simply from the group presentation (not using central extensions, etc.).  Even nicer would be an example of how software (GAP, Magma, Magnus, etc.) could automate that.
 [/INST] OK, here is the derivation, based completely on the amazing information provided by Victor Miller (who I should also thank for letting me know about kbmag).
First, some identities:
(1) From $x^3=xyxy$ we get: (a) $x^2=yxy$; (b) $xyx^{-1}=y^{-1}x$; (c) $x^{-1}yx=xy^{-1}$.
(2) From $y^5=xyxy$ we get: (a) $y^4=xyx$; (b) $x^{-1}y^3=yxy^{-1}$; (c) $y^3x^{-1}=y^{-1}xy$.
(3) From (1a) and (3b) we get $(yxy)(yxy^{-1})=(x^2)(x^{-1}y^3) = xy^3$; so $xy^2xy^{-1}=y^{-1}xy^3$.
(4) From (2b) and (1b) we get $(yxy^{-1})(xyx^{-1}) = (x^{-1}y^3)(y^{-1}x) = x^{-1}y^2x$, so that $yxy^{-1}xy=x^{-1}y^2x^2$.
(5) From (2c) we get $y^2x^{-1}y^{-1}=y^{-2}x$; squaring that yields $y(yx^{-1}yx^{-1})y^{-1}=y^{-2}xy^{-2}x$.  (1c), inverse, squared, shows this is the same as $yx^{-1}y^{-2}xy^{-1}=y^{-2}xy^{-2}x$.
(6) Similar to (5).  From (2c) we get $y^2x^{-1}=y^{-2}xy$; squaring that yields $y^2x^{-1}y^2x^{-1}=y^{-1}(y^{-1}xy^{-1}x)y$.  (1b) squared shows this is the same as $y^2x^{-1}y^2x^{-1}=y^{-1}xy^2x^{-1}y$.
OK, now consider the word $(y^{-1}xy^3)xy^{-1}xy$. From (3) this is $xy^2x(y^{-1}xy^{-1}x)y$, which from (1b) squared is $xy^2x(xy^2x^{-1})y=xy^2x^2y^2x^{-1}y$.
This word can also be written as $y^{-1}xy^2(yxy^{-1}xy)$, which from (4) is $y^{-1}xy^2(x^{-1}y^2x^2)$. So the previous two computations show
$y^2x^2y^2xy^{-1}=x^{-1}(y^{-1}xy^2x^{-1}y)yx^2$
$=x^{-1}y^2x^{-1}y^2(x^{-1}yx)x$ ...... from (6)
$=x^{-1}y^2(x^{-1}y^2x)y^{-1}x$ ....... from (1c)
$=(x^{-1}y^2x)y^{-1}xy^{-2}x$ ......... from (1c) squared
$=xy^{-1}x(y^{-2}xy^{-2}x)$ ........... from (1c) squared
$=xy^{-1}(xyx^{-1})y^{-2}xy^{-1}$ ..... from (5)
$=x(y^{-2}xy^{-2}x)y^{-1}$ .............from (1b)
$=(xyx^{-1})y^{-2}xy^{-2}$ ............ from (5)
$=y^{-1}xy^{-2}xy^{-2}$ ............... from (1b).
So $y^2x^2y^2x^{-1}y=y^{-1}xy^{-2}xy^{-2}$, or $y^2x^2y^2=y^{-1}xy^{-2}xy^{-3}x$. But
$y^{-1}xy^{-2}x(y^{-3}x)=y^{-1}xy^{-2}(xyx^{-1})y^{-1}=y^{-1}x(y^{-3}x)y^{-1}=y^{-1}(xyx^{-1})y^{-2}=y^{-2}xy^{-2}$ (through repeated application of (2b) inverse, and (1b)).
Thus $y^2x^2y^2=y^{-2}xy^{-2}$, or $y^4x^2y^4=x$, and from (2a) we get $xyx^4yx=x$, or $1=yx^4yx=(yxy)x^4=x^6$ (the second follows from $x^3$ being central and the third from (1a)).
Done! 
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