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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.

Related facts

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.