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](https://mathoverflow.net/questions/64195/when-is-a-finite-dimensional-real-or-complex-lie-group-not-a-matrix-group) 


"
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. 



Why a nilpotent Lie group must be a matrix group?
Another reference which was not yet mentioned, I think, is the article of M. Moskowitz, "Faithful Representations and a local property of Lie groups", Math. Z. $143$, 1975. 
There the question is discussed when *all* analytic groups with a given Lie algebra $\mathfrak{g}$ have a faithful linear representation. He proves among other things the following result:


**Theorem** *All* connected Lie groups $G$ with Lie algebra $\mathfrak{g}$ have a faithful linear representation if and only if $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$ and $\tilde{S}$ has a faithful representation.


Here $\tilde{G}=rad(\tilde{G})\cdot \tilde{S}$ is the Levi decomposition for the simply connected group $\tilde{G}$ with Lie algebra $\mathfrak{g}$.


**Corollary** Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If $ad(\mathfrak{g})$ has nontrivial center then there is a locally isomorphic group without any faithful representation.


