The question may be a little naive (or even appear as a duplicate) as I guess the result is well known. I saw on the other thread that
" c) A solvable Lie group G is linear iff its commutator subgroup G′ is closed, and G′ has no non-trivial compact subgroup (Thm 3.2 in Chap. XVIII) "
But I cannot find the reference book anywhere. May I ask someone to give a hint why this is true?
Background:
Try to find a non-trivial example of infranil manifold which is not a matrix Lie group. Then the speaker visiting my university claiming every nilpotent Lie group must be a matrix Lie group. Since nilpotent is defined entirely algebraically, I feel there should be a proof based on algebraic techniques. But I do not know how to prove the above statement.