$\mathbf{n}$ is nilpotent Lie algebra with $N$ being the corresponding *algebraic* Lie group. Now one neat feature of this setting is that you can take the exponential map to be identity. In other words you can define a group structure on $\mathbf{n}$ using the Campbell-Hausdorff formula. I have the following questions (they might be very easy; I just don't know):


1. If $\pi: \mathbf{n} \rightarrow \text{End}(V)$ is a representation of the nilpotent Lie algebra then does $x \in N$ acts as $\exp (\pi (x))$ on $V$ ?
2. If $\mathbf{m}$ is a subalgebra of $\mathbf{n}$ then you get a corresponding group $M \subset N$. Do we have: $N/M = \mathbf{n}/\mathbf{m}$ as sets?


**EDIT:** Victor's comments are to the point so in order to clear up the confusion I added that $N$ is algebraic.



Nilpotent Lie algebras and unipotent Lie groups
I will suppose that $\mathfrak n$, etc is finite-dimensional.


If by "the corresponding Lie group" you mean "the corresponding connected simply-connected Lie group", then what you say is correct (in spite of Victor Prostak's comment). Let $\{\mathfrak n,[,]\}$ be a Lie algebra over a field $k$ of characteristic $0$. Then the BCH formula truncates, and in particular converges. In general, BCH defines a "partial group" structure in its domain of convergence (as for "typical uses", in finite dimensions over $\mathbb R$, BCH has positive radious of convergence, and [by gluing together such patches one can prove the Lie III theorem](https://mathoverflow.net/questions/8784/why-is-lies-third-theorem-difficult)), and so for nilpotent $\mathfrak n$ it defines a group structure $\{\mathfrak n, \operatorname{BCH}\}$ on the set of points in $\mathfrak n$. In fact, when $\{\mathfrak n,[,]\}$ is nilpotent, the fact that BCH truncates means that it is polynomial, and so $\{\mathfrak n, \operatorname{BCH}\}$ is affine algebraic over $k$. In particular, when $k = \mathbb R$, then BCH defines a real algebraic group $\{\mathfrak n, \operatorname{BCH}\}$, and in particular a Lie group, and it is straightforward to check that $\operatorname{Lie}(\{\mathfrak n, \operatorname{BCH}\}) = \{\mathfrak n,[,]\}$. Since as a topological space $\mathfrak n$ is connected and simply connected, $\{\mathfrak n, \operatorname{BCH}\}$ is the connected simply-connected group corresponding to $\{\mathfrak n,[,]\}$. This I think you already know, but I thought it best to spell it out for other readers.


You ask two questions.


(1.) Yes, if I'm understanding correctly. Let $\pi : \{\mathfrak n,[,]\} \to \operatorname{End}(V)$ be a Lie algebra homomorphism, and $\dim V < \infty$. Since $\{\mathfrak n, \operatorname{BCH}\}$ is connected and simply connected, $\pi$ determines a group homomorphism $\Pi: \{\mathfrak n, \operatorname{BCH}\} \to \operatorname{GL}(V)$. Given $x\in \mathfrak n$, yes, $\Pi(x) = \exp(\pi (x))$. 


There's nothing special about unipotent groups here. Let $\mathfrak g$ be a (finite-dimensional) Lie algebra (over $\mathbb R$) and $G$ the corresponding connected simply-connected Lie group, and let $\exp : \mathfrak g \to G$ be the exponential map (in your example, $\exp: \{\mathfrak n,[,]\} \to \{\mathfrak n, \operatorname{BCH}\}$ is an isomorphism of spaces, so we call it the identity). Let $H$ be any other Lie group and $\mathfrak h = \operatorname{Lie}(H)$, and let $\exp: \mathfrak h \to H$ be the exponential map. Then there is a canonical bijection between Lie algebra homomorphism $\mathfrak g \to \mathfrak h$ and Lie group homomorphisms $G \to H$, and the bijection intertwines the two exponential maps ($\Phi\circ \exp = \exp \circ \phi$). One proof is uses the following description of $\exp$: we can identify $\mathfrak g \hookrightarrow \Gamma({\rm T}G)$ as left-invariant vector fields, so that $x\in \mathfrak g$ determines an ODE on $G$, and it turns out that this ODE has solutions for all times, and $\exp(x)$ is the image of $e\in G$ under the flow-by-time-$1$ map.


(2.) No. I think what you are asking is the following. Let $\{\mathfrak n,[,]\}$ be a (finite dimensional) nilpotent Lie algebra (over $\mathbb R$) and $\{\mathfrak m,[,]\}$ a subalgebra. Then we have connected simply-connected groups $\{\mathfrak n,\operatorname{BCH}\}$ and $\{\mathfrak m,\operatorname{BCH}\}$, and the latter is canonically a subgroup of the former. Then I think what you are asking is whether the cosets of $\{\mathfrak m,\operatorname{BCH}\}$ in $\{\mathfrak n,\operatorname{BCH}\}$ are the same as the cosets of the vector space $\{\mathfrak m,+\}$ in the vector space $\{\mathfrak n,+\}$.


Consider the three-dimensional Heisenberg algebra, which is generated by two elements $x,y$, with the condition that $[x,y]$ is central (then $x,y,[x,y]$ form a basis). This has the matrix representation:
$$ x = \left( \begin{array}{ccc} 0 & 1 & 0 \\ & 0 & 0 \\ & & 0 \end{array} \right), \quad y = \left( \begin{array}{ccc} 0 & 0 & 0 \\ & 0 & 1 \\ & & 0 \end{array} \right), \quad [x,y] = \left( \begin{array}{ccc} 0 & 0 & 1 \\ & 0 & 0 \\ & & 0 \end{array} \right)$$
Then $\operatorname{BCH}(ax,y) = ax + y + \frac a 2 [x,y]$. In particular, letting $\mathfrak n$ be the Heisenberg algebra and $\mathfrak m$ the one-dimensional subalgebra spanned by $x$, we see that the (right, say) cosets of $\{\mathfrak m,\operatorname{BCH}\}$ do not agree with the cosets of $\{\mathfrak m,+\}$: in particular, the cosets through $y$ are different (one is a line parallel to the span of $x$, the other is a line parallel to the span of $x + \frac12 [x,y]$).


Note that as Emerton points out, if $\{\mathfrak m,[,]\}$ is a Lie ideal in $\{\mathfrak n,[,]\}$, then the answer is yes. Let $y \in \mathfrak n$. Then the coset of $\{\mathfrak m,\operatorname{BCH}\}$ through $y$ consists of elements of the form $x + y + l$, where $x \in \mathfrak m$ and $l$ is a Lie polynomial with at least one $x$, and hence (since $\mathfrak m$ is an ideal), it is in $\mathfrak m$. I.e. the coset is precisely $y + \mathfrak m$.


