Show $A_5 \cong \langle x,y \mid x^5,y^2,(xy)^3\rangle$ I want to prove the following:
$$A_5 \cong \langle x,y \mid x^5,y^2,(xy)^3\rangle=:K.$$
I know that $A_5=\langle (1,2,3,4,5),(1,2)(3,4)\rangle$ and #$A_5=60$.
Let $\alpha=(1,2,3,4,5), \beta=(1,2)(3,4).$ Then $\alpha^5=\beta^2=(\alpha\beta)^3=1$ holds. So there exists an epimorphism $\phi:K\to A_5, \phi(x)=\alpha,\phi(y)=\beta$. Especially, #$K\geq 60$.
Now I want to show #$K\leq 60$ by getting the elements of $K$ in some kind of normal form. I know $y=y^{-1}, xyx=yx^{-1}y, yxyx=x^{-1}y=x^4y$. Somehow, I cannot manage to achieve the normal form. Any help is greatly appreciated!
 A: The question asks for an "ordered / structured enumeration" of the to-be $k=60=5!/2$ elements of $K$ generated by $x,y$ with the presentation $1=x^5=y^2=(xy)^3$. I would like to work as follows. We model the structure of the enumeration modeled / based on some subgroup of the to-be isomorphic group $A_5$.
So enumerate the elements of some subgroup $H<K$ of order $h|k=60$ parallel to some subgroup of order $h$ of $A_5$, then use left or right classes to complete the parallel.
I decided to pick the divisor $h=10$ of $k=60$, and the group
$$
H=\Big\langle \ y,z\ \Big\rangle\ ,\qquad z:=xyx^{-1}
\ .
$$
As a conjugate of $y$, $z$ has also order two.
Let us show that $K$ has order $10$, it is not commutative, thus dihedral, and enumerate the ten elements. Consider
$$w=yz\ .$$
Its inverse is $zy$ because the multiplication of the two elements gives $(yz)(zy)=y(z^2)y=yy=1$. The order of $w$ is five, let us show this.
$$
\begin{aligned}
xyx &= yx^{-1}y \qquad\text{(recall)}\\
yxy &= x^{-1}yx^{-1} \qquad\text{(recall)}\\[2mm]
w^5 
&=yz\ yz\ yz\ yz\ yz\\
&=yxyx^{-1}\  yxyx^{-1}\  yxyx^{-1}\  yxyx^{-1}\  yxyx^{-1}\  \\
&=\color{blue}{yxy}\ x^{-1}\  \color{blue}{yxy}\ x^{-1}\  \color{blue}{yxy}\ x^{-1}\  \color{blue}{yxy}\ x^{-1}\  \color{blue}{yxy}\ x^{-1}\  \\
&=
\color{blue}{x^{-1}yx^{-1}}\ x^{-1}\  
\color{blue}{x^{-1}yx^{-1}}\ x^{-1}\  
\color{blue}{x^{-1}yx^{-1}}\ x^{-1}\  
\color{blue}{x^{-1}yx^{-1}}\ x^{-1}\  
\color{blue}{x^{-1}yx^{-1}}\ x^{-1}\\
&=
x^{-1}\ \color{maroon}{yx^2y}\ x^2\ \color{maroon}{yx^2y}\ x^2\ yx^{-2}
\\
&=
x^{-1}\ \color{maroon}{(yxy)(yxy)}\ x^2\ \color{maroon}{(yxy)(yxy)}\ x^2\ yx^{-2}
\\
&=
x^{-1}\ \color{maroon}{(x^{-1}yx^{-1})(x^{-1}yx^{-1})}\ x^2\ \color{maroon}{(x^{-1}yx^{-1})(x^{-1}yx^{-1})}\ x^2\ yx^{-2}
\\
&=
x^{-1}\ \color{maroon}{(x^{-1}yx^{-1})(x^{-1}\color{red}{y})}\ \color{maroon}{(\color{red}{y}x^{-1})(x^{-1}y)}\ x\ yx^{-2}
\\
&=
x^{-2}yx^{-2}\color{red}{y\ y}x^{-2}\ \color{blue}{yxy}\ x^{-2}
\\
&=
x^{-2}yx^{-2}\ x^{-2}\ \color{blue}{x^{-1}yx^{-1}}\ x^{-2}
\\
&=
x^{-2}y\ \color{red}{x^{-5}}\ \color{blue}{yx^{-1}}\ x^{-2}
\\
&=
x^{-2}y\ \color{blue}{yx^{-1}}\ x^{-2}
\\
&=
x^{-5}
\\
&=1\ .
\end{aligned}
$$
So $H$ contains as subgroup $\langle w\rangle=\{1,w,w^2,w^3, w^4\}$, and because of $zwz=z(yz)z=zy=w^{-1}$, we see that $K=\langle y,z\rangle=\langle y,w\rangle$ is the dihedral group with ten elements, explicitly written for instance as
$$
\begin{aligned}
H
&=\{\ 1,w,w^2,w^3,w^4\ ;\ y,yw,yw^2,yw^3,yw^4\ \}\\
&=\{\ 1,w,w^2,w^3,w^4\ ;\ y,wy,w^2y,w^3y,w^4y\ \}
\ .
\end{aligned}
$$
Consider now the set $R$ (a posteriori of representatives for cosets)
$$
R =\{\ 1,x,x^2,x^3, x^4;\ x^3y\ \}
\ .
$$
Lemma:
A full enumeration of the elements in $K$ is:
$$K=HR:=\{\ hr\ :\ h\in H\ ,\ r\in R\ \}\ .$$
Proof of the Lemma:
It is enough to show that multiplication from the left with $x$, and with $y$ invariates the finite set $HR$ with $60$ elements. It is clear that $yHR=HR$ since $yH=H$. Let us consider now multiplication with $x$. It is enough to see the following relations in $K$:
$$
\begin{aligned}
x\cdot 1 &=1\cdot x\in HR\ ,\\
x\cdot y &=y\ (y\ xyx^{-1})\ x=y\ (yz)\ x =yw\ x\in HR\ ,
\\
yxy=yxy^{-1} &=wx\ ,
\\
xyx^{-1} &=yw\ ,
\\[2mm]
x\cdot w
&=xyz=xy\ xy\ x^{-1}=(xy)^2\ x^{-1}=(xy)^{-1}\ x^{-1}=yx^4\ x^{-1}
=y\cdot x^3\in HR\ ,
\\
xwx^{-1} &=yx^2\ ,
\\
x\cdot w^2
&=(xw^2x^{-1})x=(xwx^{-1})^2x=(yx^2)^2 x=yx^2yx^3
\\
&=(yxy)^2x^3
=(wx)^2x^3=wxwx^4=w(xwx^{-1})=wy\cdot x^2\in HR\ ,
\\
x\cdot w^3
&=(xw^2)w=wyx^2\ w=wyx\ xw=wyx\ yx^3=wy(xy)x^3\\
&=wy(ywx)x^3=w^2\cdot x^4\in HR\ ,
\\
x\cdot w^4
&=xw^{-1}=x(yxyx^{-1})^{-1}=x(xyx^{-1}y)=x^2yx^4y=x(xyx)x^3y=x(yx^{-1}y)x^3y
\\
&=(xyx^{-1}y)x^3y=w^{-1}x^3y
=w^4\cdot  x^3y
\in HR\ ,
\\[2mm]
x\cdot yw^k
&=
(xy)\ w^k =yw\ x\ w^k\in yw\ HR\subseteq H\ HR=HR
\ .
\end{aligned}
$$
$\square$
The above shows one possibility to enumerate the elements, it is not "the" normal form, but shows how some structured enumeration can be achieved, based on a parallel "splitting" the structure of $A_5$. Since $A_5$ is simple, there is no normal subgroup that can be used, but some choice of a subgroup of a suitable order (e.g. as close as possible to $\sqrt {|A_5|} =\sqrt {60}$), and corresponding classes may be a good plan.
