Right Angled Artin Group Reference request 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?
 A: 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. 
A: 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 $\langle x, x^g \rangle \cong F_2$.
