Lazard Lie property is not subgroup-closed

Statement
It is possible to have a Lazard Lie group $$G$$ and a subgroup $$H$$ of $$G$$ that is not a Lazard Lie group.

Similar facts

 * LCS-Lazard Lie property is not subgroup-closed
 * Baer Lie property is not subgroup-closed

Opposite facts

 * Powering-invariant subgroup of Lazard Lie group is Lazard Lie group

Proof
Consider the example $$G = UT(3,\mathbb{Q})$$ and $$H = UT(3,\mathbb{Z})$$. $$G$$ is a Lazard Lie group (in fact, a Baer Lie group). $$H$$ is a group of class exactly two that is not 2-powered, hence, it is not a Lazard Lie group.