An $n$-dimensional complex manifold $M$, endowed with an Hermitian metric $g$, is K"ahler if and only if the holonomy group of $g$ is contained in $U(n)$. If $g$ is Hermitian but not K"ahler do we necessarily get a reduction of the holonomy group.

A manifold is called hypercomplex if it admits three integrable complex structures $I,J,$ and $K$ which together give a representation of the quaternions $\mathbb{H}$. Each complex structure will admit three Hermitian metrics. Can we conclude any reduction of their holonomy groups?

An $n$-dimensional complex manifold $M$, endowed with an Hermitian metric $g$, is Kähler if and only if the holonomy group of $g$ is contained in $U(n)$. If $g$ is Hermitian but not Kähler do we necessarily get a reduction of the holonomy group.

EDIT: To clarify, I mean holonomy group with respect to the Chern connection, i.e. the second connection specified by Robert below.

A manifold is called hypercomplex if it admits three integrable complex structures $I,J,$ and $K$ which together give a representation of the quaternions $\mathbb{H}$. Each complex structure will admit three Hermitian metrics. Can we conclude any reduction of their holonomy groups?