Lazard Lie property is not subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., Lazard Lie group) not satisfying a group metaproperty (i.e., subgroup-closed group property).
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Statement

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

Related facts

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

Further information: unitriangular matrix group:UT(3,Q), unitriangular matrix group:UT(3,Z)

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.