TITLE: Deriving a relation in a group based on a presentation
QUESTION [12 upvotes]: 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.

REPLY [3 votes]: It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $Ext(C_{30},G)$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$
$$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = u^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ 
Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too. 
The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, 
Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

REPLY [3 votes]: This is a very basic answer to the last part of the question. One can derive the relation $x^6=1$ in gap and magma and even identify the group in this case. As a word of caution, these methods may break down depending on the automation one has in mind.
In gap:
 gap> F:= FreeGroup(2); x:=F.1; y:=F.2;
 <free group on the generators [ f1, f2 ]>
 f1
 f2
 gap> G:= F/[x^3*y^-5,x^-3*(y*x)^2];    
 <fp group on the generators [ f1, f2 ]>
 gap> Order(Subgroup(G,[G.1]));        
 6

To identify the whole group:
gap> Order(G); # note this line is not strictly necessary
120
gap> StructureDescription(G);
"SL(2,5)"

In magma, we can do the same operations:
 > G<x,y>:=Group<x,y|x^3=y^5=(y*x)^2>; 
 > Order(sub<G|x>);
 6
 > Order(G); # again this line is not strictly necessary
 120
 > IdentifyGroup(G);
 <120, 5> 

From here, you look up the group as the 5th group of order 120 in the small group data base (see http://magma.maths.usyd.edu.au/magma/handbook/text/703). In the alternative, you could put in your favorite presentation of $SL(2,5)$ and check that is it also group "<120, 5>."