Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the completed free Lie algebra on the same number of generators as $G$, by the (formal) logarithm of the defining relations of $G$ (see edit below).

$G$ is called 1-formal if its Malcev Lie algebra is isomorphic as a filtered Lie algebra to the degree completion of a finitely presented Lie algebra with quadratic relations. (If $X$ is a path-connected topological space such that $G=\pi_1(X)$, this is equivalent to say that the Malcev Lie algebra of $G$ is isomorphic to the holonomy Lie algebra of $X$)

It is quite well known that fundamental groups of complementary of hyperplane arrangements in $\mathbb{C}^n$ (e.g. pure braid groups) are 1-formal.

On the one hand, an important fact in the study of these groups is that they are (under some conditions on the underlying arrangement) iterated "almost-direct product" of free groups, and free groups are themselves 1-formal. An almost direct product is a semi-direct product $H\rtimes K$ for which the action of $K$ on the abelianization of $H$ is trivial. On the other hand, according to this interesting survey http://www.arxiv.com/abs/0903.2307 the direct product of two 1-formal groups is again 1-formal, even if no proof is given (it's probably easy..)

So it is temptating to ask:

is an almost-direct product of 1-formal groups again 1-formal ?

**Edit:** Some details about the construction and its relation with almost direct product. If $G$ is an abelian group, then one can take the tensor product $G \otimes_{\mathbb{Z}} \mathbb{Q}$ leading to a uniquely divisible abelian group. The Malcev construction extends this to any nilpotent group, leading to a uniquely divisible nilpotent group. Let $G^{(0)}=G$, $G^{(n+1)}=[G^{(n)},G]$ be the lower central serie, then the quotients $G/G^{(n)}$ are nilpotent by construction.

The Malcev completion of $G$ is the inverse limit of the $(G/G^{(n)}) \otimes_{\mathbb{Z}} \mathbb{Q}$. It is a pro-unipotent group, hence it has a (pro-nilpotent) Lie algebra $\mathfrak g$. On the other hand, one can define a Lie algebra by $$gr\ G=\mathbb{Q}\otimes_{\mathbb{Z}} \bigoplus G^{(n)}/G^{(n+1)}$$ the bracket being induced by the commutator in $G$. It as a graded Lie algebra, and in fact it is the associated graded of the filtered Lie algebra $\mathfrak g$.

A group is called 1-formal if there is an isomorphism of filtered Lie algebras $gr\ G \cong \mathfrak g$. Now, the point is that almost-direct product behave well with respect to the lower central series. If $G=G_1\rtimes G_2$ is an almost-direct product, then it seems to me that a result of Falk and Randell implies that we have $$G^{(n)}=G_1^{(n)}\rtimes G_2^{(n)}$$ and $$gr\ G = gr\ G_1 \rtimes gr\ G_2$$.