# 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!

• Does this answer your question? Group presentation of $A_5$ with two generators Nov 30, 2022 at 17:45
• @AnneBauval thank you, I have seen this question but the proof does not bring the elements on a normal form and I want to know how / if this is possible.. Nov 30, 2022 at 17:48
• May be there is no "normal form" but this is harder (to formulate and prove) than showing $|K|\le60.$ Nov 30, 2022 at 17:53
• You can use the fact that $A_5$ is the group of rotational symmetries of a dodecahedron. You can then sketch out an "$A_5$-symmetric" embedding into the dodecahedron of the Cayley 2-complex of $K$, and check that it has 60 vertices (5 in each face of the dodecahedron). Nov 30, 2022 at 18:13
• A with any finite group, there is a regular normal form consisting of a shortest word over the alphabet $\{x,x^{-1},y\}$ for each group element, using a lexicographical order to break ties. But that is not particularly helpful in computing the order of the group unless you are doing it by computer. In this example the normal form is accepted by an automaton with $25$ states. Nov 30, 2022 at 18:49

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 $$H of order $$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.