# Quotient of an Artin group is an Artin group

I'm working on a problem about Artin groups, and to simplify this problem I want to take a quotient that allow us to go to an easier Artin group, but I'm not sure if the quotient is well defined. This is the statement of my problem:

Let $$A_\Gamma$$ be an Artin group, where $$\Gamma$$ is a complete graph, $$f:A_\Gamma\to\mathbb{Z}$$ a group homomorphism and $$a\in V(\Gamma)$$ such that $$f(a)=1$$. The artin group is generated by the vertices of the graph, and there are two types of generators: the $$v\in V(\Gamma)$$ such that $$f(v)=0$$ and the $$v\in V(\Gamma)$$ such that $$f(v)\neq0$$. My objective is to define a normal subgroup $$N\trianglelefteq A_\Gamma$$ such that $$A_\Gamma/N\cong A_{\Gamma'}$$ where $$\Gamma'\subset\Gamma$$ is the induced subgraph by the $$v\in V(\Gamma)$$ such that $$f(v)=0$$ and $$a$$, i.e. we want a new Artin group given by colapsing all vertices with non-zero image to $$a$$. To do so, I wanted to define $$N$$ as the normal subgroup generated by the set: $$\lbrace v^{-1}a^{f(v)} \mid v\in V(\Gamma),f(v)\neq0\rbrace$$ (Remark: I $$N\leq\ker(f)$$ if possible, that is why I need to define those ''weird'' generators instead of $$v^{-1}a$$) It is obvious that in the quotient $$A_\Gamma/N$$ all the $$v\in V(\Gamma)$$ with $$f(v)\neq 0$$ are identified with some power of $$a$$, which allow us to get rid of those generators. However, I can't see if this quotient is indeed the Artin group I want or sth different.

Any hint will be thanked.

Edit: Inspired by @MoisheKohan I was able to see that I can reduce the problem to the right-angled case by first taking a quotient with the normal subgroup generated by the commutators of the generators. In this situation, since $$\Gamma$$ is a complete graph $$A_\Gamma\cong\mathbb{Z}^n$$, where $$n$$ is the number of generators of $$\mathbb{Z}$$, so the problem should be easier.

• You did not specify what type of Artin groups you are working with: do you assume right-angled? Apr 17, 2023 at 22:22
• @MoisheKohan I wasn't working with right-angled groups, but I managed to simplify the problem to this case. Hope it helps to solve the problem. Apr 18, 2023 at 9:47
• What's with the $a^{f(v)}$ thing? You say you want to identify all the "living" $v$ with $a$, do you mean with a power of $a$? Or should your set of normal generators for $N$ consist of $v^{-1}a$ rather than $v^{-1}a^{f(v)}$? Apr 18, 2023 at 10:50
• @MattZaremsky I wanted to do this because I want $N\leq \ker(f)$. And yes, then I have identified $v$ with a power of $a$, but that is not a problem for my purposes, since this is enough to get rid of the generator $v$. I'll edit the question. Apr 18, 2023 at 11:02

If I'm understanding all this correctly, the answer is "no". Take three vertices, $$a$$, $$b$$, and $$c$$. Take an edge labeled 4 from $$a$$ to $$b$$, an edge labeled 4 from $$a$$ to $$c$$, and an edge labeled 2 from $$b$$ to $$c$$. Take $$f$$ to be $$f(a)=f(b)=1$$, $$f(c)=0$$. So now you're modding out that $$a$$ and $$b$$ get identified. In this quotient, $$a$$ and $$c$$ suddenly commute, which is not what the induced subgraph $$\Gamma'$$ wants (it's $$a$$ and $$c$$ with an edge labeled 4). If you want to phrase it in terms of quotient maps instead of normal subgroups, the problem is that the quotient map isn't well defined: you have that $$b$$ and $$c$$ commute, but their images are $$a$$ and $$c$$, which don't commute.
• Thanks for your time! However, one of the hypothesis I haveis that $\Gamma$ is a complete graph, so when passing to the RAAG case this is equivalent to say that $A_\Gamma=\mathbb{Z}^n$. Since I've added the RAAG hypothesis later I forgot to mention that, sorry. Apr 18, 2023 at 11:13
• Well, now I think about this it seems trivial that we can take a quotient of $\mathbb{Z}^n$ to get $\mathbb{Z}^m$ getting rid of the generators we want, am I right? Apr 18, 2023 at 11:23