# Presentation $\langle x,y \mid x^3=y^3=(xy)^3=1\rangle\cong\langle t\rangle\ltimes A$

Hi: This question has already been answered here: Show $\langle x,y|x^3=y^3=(xy)^3=1\rangle$ is isomophic to $A\rtimes\langle t\rangle$, where $t^3=1$ and $A=\langle a\rangle\times\langle b\rangle$.

However I don't understand the answer so I post.

Let $$G=\langle x,y \mid x^3=y^3=(xy)^3=1\rangle$$. Prove that $$G$$ is isomorphic to the semidirect product of $$A$$ by $$\langle t\rangle$$ where $$t^3=1$$ and $$A=\langle a\rangle \times \langle b\rangle$$ is the direct product of two infinite cyclic groups, the action of $$t$$ being $$a^t=b, b^t=a^{-1}b^{-1}$$.

[Hint: Prove that $$\langle xyx,x^2y\rangle$$ is a normal abelian subgroup.]

First I'll prove that the semidirect product is generated by two elements satisfying the same relations as those of $$G$$. Let $$L$$ be the semidirect product of $$A$$ by $$\langle t\rangle,$$ and let $$u=(t,1), v=(t,ab)$$. Then $$uv=(t^2, 1^t ab)=(t^{-1},ab)$$, and \begin{align}\tag{1}(uv)^2&=(t,(ab)^{t^{-1}}ab)=(t,(ab)^{t^2}ab)\\&=(t,[a^t b^t]^t ab)=(t,(ba^{-1}b^{-1})^t ab)\\&=(t,(a^{-1})^t ab)=(t,(a^t)^{-1}ab)\\&=(t,b^{-1}ab)=(t,a).\end{align}

Computing also shows $$v^2=(t^{-1},b), (t,1)^{-1}(t,a)=(t^{-1},1)(t,a)=(1,a)$$ and $$(t,1)(t^{-1},b)=(1,b)$$. So $$u, v$$ generate $$\langle (t,1), (1,a), (1,b)\rangle$$. But these generate $$L$$. That is, $$L=\langle u,v\rangle$$.

Also it is easy to see that $$u^3=1$$ and $$v^3=1$$. And by (1) $$(uv)^3= (uv)^2uv=(t,a)(t^{-1},ab)=(1,a^{t^2}ab)=(1,b^t ab)=(1,a^{-1}b^{-1}ab)=(1,1)$$ because A is abelian, proving the assertion.

As every relation in $$G$$ is a relation in $$L$$, by von Dyck's theorem there is an epimorphism $$\theta: G \to L$$. If I could prove ker $$\theta =1$$ then $$G$$ would be isomorphic to $$L$$. But how do I prove it?

EDIT: I forgot the hint. There should be some way to use it. Let the subgroup given in the hint be $$H$$. Suppose I prove $$A\simeq H$$. Then I have, on the one hand, $$|L|=3 |A|$$ and on the other hand, if I prove $$|G/H|=3, |G|=3 |H|$$ and so $$|G|=|L|$$. Given that $$G$$ and $$L$$ are both generated by two elements satisfying the same relations, I think one can infer $$G\simeq L$$. So one thing I could do is to try to prove $$|G/H|=3$$. I will see if I can prove it and will let you know.

• You can do this directly by modifying the presentation. Letting $a=xyx, b=x^2y, t=xy$. Then $t^3=1$. From $bt^{-1}=x=t^{-1}a$ it follows $a^t=b$. From $1=x^3=(t^{-1}a)^3$ it follows $t^{-1}bt=b^{-1}a^{-1}$. Finally, $y^{-1}=ta$, and from $1=y^{-3}=(ta)^3$, it follows $[b,a]=1$. Thus your group also has the presentation $$\langle a,b,t\mid t^3=[b,a]=1, a^t=b, b^t=b^{-1}a^{-1}\rangle$$ – Steve D Jan 10 '20 at 23:43
• Your argument in the edit is not good: $A$ is an infinite group, so $|L| = 3|A|$ isn't meaningful in the way that it would be if $|A|$ were finite. – Rylee Lyman Jan 11 '20 at 3:29
• My argument in the comments above can be formalized, since it's just doing Tietze transformations. Then you would have to argue the presentation I ended with is really the semidirect product you're after. Another way is to use that presentation to inform an internal decomposition as a product. So you have $N=\langle xyx, x^2y\rangle$, and you need to show that's abelian and normal. Then you can show $G/N$ is generated by the image of $xy$. Then if $K=\langle xy\rangle$, we'll have a semidirect decomposition as long as $xy\not\in N$, which would imply $G=N$. Look for a nonabelian quotient. – Steve D Jan 11 '20 at 5:39

As @SteveD mentions in the comments, the presentation you implicitly have for $$L$$ can be transformed into a presentation for $$G$$ and vice versa.

More formally, you could apply your argument in reverse to define an epimorphism $$L\to G$$ using von Dyck's theorem. Then it should be possible to use the concrete description of the epimorphisms that you have calculated to conclude that the epimorphisms are in fact inverses, and thus isomorphisms.

Let's talk through what you have shown. We have implicitly talked of $$L$$ as having the following standard presentation

$$L = (\mathbb{Z}\times\mathbb{Z})\rtimes\mathbb Z/3\mathbb{Z} = \langle a,b,t \mid [a,b] = t^3 = 1, a^t = b, b^t = a^{-1}b^{-1} \rangle .$$

You found $$u$$ and $$v$$ in $$L$$, showed that $$\langle u,v \rangle = L$$, and that $$u$$ and $$v$$ satisfy the relations $$u^3 = v^3 = (uv)^3 = 1$$.

In this case, I would like to claim that von Dyck's theorem implies that there exists an epimorphism $$G\to L$$ defined by $$x \mapsto u$$, $$y\mapsto v$$. Indeed, pick your favorite two-element set, say $$\{g,h\}$$, and let $$F$$ be the free group on $$\{g,h\}$$. By the universal property of the free group, the functions

$$\phi \begin{cases} g \mapsto x \\ h \mapsto y \end{cases}\qquad \psi\begin{cases} g \mapsto u \\ h \mapsto v \end{cases}$$

define homomorphisms $$\Phi\colon F \to G$$ and $$\Psi\colon F \to L$$, respectively. The statements that $$\{x,y\}$$ and $$\{u,v\}$$ generate $$G$$ and $$L$$, respectively implies that $$\Phi$$ and $$\Psi$$ are presentations. The work you did in the OP allows us to apply von Dyck's theorem to say that the map $$f\colon G \to L$$ defined by $$x \mapsto u$$ and $$y \mapsto v$$ is a well-defined epimorphism.

My above suggestion is that you should argue in a similar way that there exists a similar generating set of $$G$$ satisfying the relations for $$L$$, so that von Dyck's theorem gives you an epimorphism $$L \to G$$. I will leave this to you. What's more, if your description of the action of the homomorphism $$L \to G$$ on the set $$\{a,b,t\}$$ is clear enough, you should be able to prove that each double composition is the identity homomorphism, perhaps by showing that the action of each double composition on the appropriate set of generators is the identity.

• Actually I was wrong in applying von Dyck. Von Dyck's theorem, at least in the book I am studying, speaks about two presentations. In order to apply it I would have to show $L$ has the presentation $L=\langle u,v \mid u^3=v^3=(uv)^3=1\rangle$. But I do not know if these three relations completely define the semidirect product $L$. There may be other necessary relations. – stf91 Jan 11 '20 at 0:59
• Ahhh, I see. I didn't know the argument I was thinking of had the name von Dyck's theorem until today. I will expand my answer. – Rylee Lyman Jan 11 '20 at 1:01
• Let G and H be groups with presentations e: F -4 G and «5: F -4 H such that each relator of e is also a relator of «5. Then the function f' f-+ f~ is a well-defined epimorphism from G to H. – stf91 Jan 11 '20 at 1:05
• In my book von Dyck's theorem is this: Let G and H be groups with presentations $\epsilon: F \to G$ and $\delta: F \to H$ such that each relator of $\epsilon$ is also a relator of $\delta$. Then the function $f^\epsilon \mapsto f^\delta$ is a well-defined epimorphism from G to H. – stf91 Jan 11 '20 at 1:12
• @stf91 ahhh, good. von Dyck's theorem does apply. I wrote a little about how to see that. – Rylee Lyman Jan 11 '20 at 3:09

I will prove $$G\simeq L$$ in a trivial way. Von Dyck's theorem applied in both directions gives me two epimorphisms, $$\theta: G \to L, x\mapsto u, y\mapsto v$$ and $$\rho: L\to G, u\mapsto x, v\mapsto y$$. I'll show $$\theta \rho = \rho \theta=1$$ with an example. Let $$xyx \in G$$. Then $$(xyx)\theta \rho=(uvu)\rho=xyx$$ and, as $$L=\langle u,v\rangle$$, let $$uv^{-1}u \in L$$, so $$(uv^{-1}u)(\rho \theta)=(xy^{-1}x)\theta=uv^{-1}u$$. Therefor $$\theta$$ is a bijection and, so, an isomorphism.

But what about the author's hint? There must be a mistake here. What about this?: Let $$\Delta=\{x^3,y^3,(xy)^3\}, R=\Delta^F$$, the normal closure of $$\Delta$$. If $$N=\langle xyx,x^2y\rangle$$ I think $$R=N$$. Let's assume this is true. If $$xy\in N$$ then $$(xy)(xyx)=(xyx)(xy)$$ because $$N$$ is abelian. But this gives $$yx=xy$$ and, so, there would be a redundant relation in $$G$$. So $$xy\notin N$$. Now $$(xy)^3 \in R$$ according to the presentation for $$G$$. So $$(Nxy)^3=1$$ and $$[G:N]=3$$ and $$G/N=\langle Nxy\rangle$$.

Let $$K=\langle xy\rangle$$. Let $$g\in G$$. Then $$Ng\in \langle Nxy\rangle, Ng=N(xy)^i, g(xy)^{-i}=n, g=n(xy)^i$$. So $$G=NK$$. Also $$N\cap K=1$$ obviously and $$G=K\ltimes N$$. Now let $$t=xy, a=xyx, b=x^2y$$. Then $$t^{-1}at=y^{-1}x^{-1}xyxxy=x^{-1}y=x^2y=b$$. That is, $$a^t=b$$. But Steve D has already proved G has also the presentation $$\langle a,b,t\mid t^3=[a,b]=1,a^t=b, b^t=a^{-1}b^{-1}\rangle$$ which seemingly is $$L$$, the semidirect product given in the problem statement. As there is no relation of the form $$a^n=1$$, I think $$\langle a\rangle$$ is infinite cyclic. Idem $$\langle b\rangle$$. I would have to prove they intersect trivially. For instance suppose $$xyx\in \langle a\rangle\cap \langle b\rangle$$. Say $$xyx=x^{-1}y$$. Then $$xyxy^{-1}x=1$$. But no nontrivial reduced word in $$\{x,y\}$$ is equal to one because $$F$$ is free on $$\{x,y\}$$. I think now the proof is complete.

• Unfortunately, there's a lot wrong in this answer. For example, just checking "an example" in the first paragraph is not enough to prove your claim that the two maps are inverses. – Steve D Jan 13 '20 at 17:20