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The following should be true: every normal subgroup of a non-Abelian Right Angled Artin Group should contain a free group on two generators. Is there a standard reference one can cite for this?

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$\mathbf{Z}\times\{1\}\subset\mathbf{Z}\times F_2$ gives a counterexample. –  Yves Cornulier Oct 4 '12 at 19:21
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Hmm. Maybe the right question should be: what is the right statement :( –  Igor Rivin Oct 4 '12 at 19:29
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Probably the starting point should be Sageev and Wise, [The Tits alternative for CAT(0) cubical complexes][1]., Bull. London Math. Soc. 37 (2005), no. 5, 706–710 So the question is: when are there non-solvable normal subgroups? [1]: ams.org/mathscinet-getitem?mr=2164832 –  Lior Silberman Oct 4 '12 at 19:58
    
@Lior: actually the question is when are there any SOLVABLE normal subgroups (I am saying never, except in trivial examples, such as direct products with abelian groups). –  Igor Rivin Oct 4 '12 at 20:08
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Igor, correct question, I think, is about center-less RAAGs. Proof should follow from Sageev-Wise with a bit of fiddling. –  Misha Oct 4 '12 at 22:38
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2 Answers

up vote 10 down vote accepted

Igor, I assume the question is about centerless RAAGs. Then it follows from the classical result of A. Baudisch (see MR0634562): every 2-generated subgroup of a RAAG is abelian or $F_2$.

Indeed if $N$ is a non-trivial normal subgroup in a RAAG $G$, take any $x\in N$. Since $G$ is centerless there exists $g\in G$ that does not commute with $x$. Then $\langle x,g\rangle\cong F_2$ and hence $\langle x, x^g \rangle \cong F_2$.

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@Denis: thanks much! The argument you give actually seems to give more than I asked for: a normal subgroup is either central or contains an $F_2.$ –  Igor Rivin Oct 5 '12 at 23:38
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A right-angled Artin group $G$ is the fundamental group of the compact non-positively curved cube complex called the Salvetti complex $X$. Thus $\tilde{X}$ is a CAT(0) space, and the action of $G$ on $\tilde{X}$ is proper and cocompact.

Thus, by the Solvable Subgroup Theorem, every virtually solvable subgroup $H< G$ contains a finitely generated abelian subgroup of finite index. Moreover, every element in $G$ will be semi-simple by Prop. 6.10 (2) from Bridson-Haefliger. By Corollary 7.2, there is an $H$-invariant closed convex subspace of $\tilde{X}$ isometric to a product $Y\times \mathbb{E}^n$, so that $H$ acts as the identity on $Y$ and acts cocompactly on the $\mathbb{E}^n$ factor. Any isometry of $\tilde{X}$ which normalizes $H$ preserves $Y\times \mathbb{E}^n$ and its splitting. We conclude that $\tilde{X}=Y\times\mathbb{E}^n$ when $H$ is normal in $G$.

Now, I think for a CAT(0) cube complex $\tilde{X}$, the factorization $Y\times \mathbb{E}^n$ should actually be a product of cube complexes. This isn't quite right, since in the group $\mathbb{Z}^n$ there are clearly $\mathbb{R}^k$ factors of $\mathbb{R}^n$ which are not a factor of the cube complex structure. But I think if $H$ is assumed to be a virtually solvable normal subgroup of $G$ and maximal with respect to this property, then $\tilde{X}=Y\times \mathbb{E}^n$ should be a product of cube complexes. I think it should be possible to prove this by analyzing links of vertices of the Salvetti complex, and how the foliation by $\mathbb{E}^n$'s passes through them.

But this implies that $G=G_1\times \mathbb{Z}^n$ where $G_1$ is a right-angled Artin group, since each edge of a cube of the Salvetti complex corresponds to a generator, so a cube complex factor isometric to $\mathbb{R}^n$ will give generators commuting with every other generator.

As observed in one of the comments, $G$ satisfies the Tits alternative since it is linear, so a normal subgroup is either virtually solvable, and then $G$ splits off a $\mathbb{Z}^n$ factor, or the normal subgroup is not solvable and therefore must contain a free subgroup. However, one would need to fill in the above missing step to finish this argument.

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