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.

Simply connectednilpotent Lie groups are linear; $\begin{pmatrix}1&\mathbf R&\mathbf R\\0&1&\mathbf R\\0&0&1\end{pmatrix}/\begin{pmatrix}1&0&\mathbf Z\\0&1&0\\0&0&1\end{pmatrix}$ isn't. See Ado's theorem. $\endgroup$ – Francois Ziegler Oct 3 '14 at 2:24