[proofplan] Let $q:G\to G/N$ be the quotient homomorphism, and study its differential at the identity. The standard quotient Lie group structure makes $q$ a smooth submersion, so $dq_e:\mathfrak g\to T_{eN}(G/N)$ is surjective. Its kernel is the tangent space to the identity fiber $q^{-1}(eN)=N$, hence equals $\mathfrak n$. Since the [differential of a Lie group homomorphism preserves Lie brackets](/theorems/8803), this kernel is an ideal, and the vector-space quotient construction gives the desired [Lie algebra](/page/Lie%20Algebra) isomorphism. [/proofplan] [step:Pass to the differential of the quotient homomorphism] Let $q:G\to G/N$ be the quotient map, so $q(g)=gN$ for every $g\in G$. Since $N$ is normal in $G$, the quotient $G/N$ is a group and $q$ is a [group homomorphism](/page/Group%20Homomorphism). By the hypothesis that $G/N$ has its standard quotient Lie group structure, $q$ is a smooth submersion. Define the [linear map](/page/Linear%20Map) $dq_e:T_eG\to T_{q(e)}(G/N)$ to be the differential of $q$ at $e$. Since $q(e)=eN$, and since $\mathfrak g=T_eG$ and $\operatorname{Lie}(G/N)=T_{eN}(G/N)$, this is a linear map $dq_e:\mathfrak g\to \operatorname{Lie}(G/N)$. Because $q$ is a submersion, the differential $dq_e$ is surjective. [/step] [step:Identify the kernel with the Lie algebra of $N$] The identity fiber of $q$ is $q^{-1}(eN)=N$. Since $q$ is a submersion and $N=q^{-1}(eN)$ is the fiber through $e$, the tangent space to the fiber at $e$ is the kernel of the differential at $e$: $T_eN=\ker(dq_e)$. Using the definition $\mathfrak n=T_eN$, we obtain $\ker(dq_e)=\mathfrak n$. [guided] The goal of this step is to replace the geometric subgroup $N$ by its infinitesimal version $\mathfrak n$. The quotient map $q:G\to G/N$ sends an element $g\in G$ to the coset $gN$. Therefore an element $g\in G$ maps to the identity coset $eN$ exactly when $gN=eN$, which is equivalent to $g\in N$. Hence the fiber of $q$ over the identity element of the quotient is precisely $q^{-1}(eN)=N$. Now use the submersion property in the standard quotient Lie group structure. For a smooth submersion, the tangent space at a point to a fiber is the kernel of the differential at that point. Applying this at the point $e\in G$ and to the fiber $q^{-1}(eN)=N$ gives $T_eN=\ker(dq_e)$. By definition of the Lie algebra of the Lie subgroup $N$, we have $\mathfrak n=T_eN$. Substituting this into the preceding identity gives $\ker(dq_e)=\mathfrak n$. This is the infinitesimal form of the statement that the quotient map collapses exactly the subgroup $N$: at the identity, the directions killed by the quotient map are exactly the tangent directions lying inside $N$. [/guided] [/step] [step:Use bracket preservation to prove that $\mathfrak n$ is an ideal] By [citetheorem:8803], the differential of the Lie group homomorphism $q:G\to G/N$ is a Lie algebra homomorphism. Thus for all $X,Y\in\mathfrak g$, $dq_e([X,Y])=[dq_e(X),dq_e(Y)]$. Let $X\in\mathfrak g$ and let $Z\in\mathfrak n$. Since $\mathfrak n=\ker(dq_e)$, we have $dq_e(Z)=0$. Therefore $dq_e([X,Z])=[dq_e(X),dq_e(Z)]=[dq_e(X),0]=0$. Thus $[X,Z]\in\ker(dq_e)=\mathfrak n$. Since this holds for every $X\in\mathfrak g$ and every $Z\in\mathfrak n$, the subspace $\mathfrak n$ is an ideal in $\mathfrak g$. [/step] [step:Descend the differential to the quotient Lie algebra] Define $\overline{dq}_e:\mathfrak g/\mathfrak n\to \operatorname{Lie}(G/N)$ by $\overline{dq}_e(X+\mathfrak n)=dq_e(X)$ for every $X\in\mathfrak g$. This map is well-defined because if $X+\mathfrak n=Y+\mathfrak n$, then $X-Y\in\mathfrak n=\ker(dq_e)$, so $dq_e(X)=dq_e(Y)$. The map $\overline{dq}_e$ is linear by linearity of $dq_e$. It is surjective because $dq_e$ is surjective. It is injective because $\overline{dq}_e(X+\mathfrak n)=0$ implies $dq_e(X)=0$, hence $X\in\ker(dq_e)=\mathfrak n$, and therefore $X+\mathfrak n=\mathfrak n$ is the zero coset. Finally, since $dq_e$ preserves Lie brackets and $\mathfrak n$ is an ideal, the quotient bracket on $\mathfrak g/\mathfrak n$ is well-defined and satisfies $\overline{dq}_e([X+\mathfrak n,Y+\mathfrak n])=\overline{dq}_e([X,Y]+\mathfrak n)=dq_e([X,Y])=[dq_e(X),dq_e(Y)]$ for all $X,Y\in\mathfrak g$. Hence $\overline{dq}_e$ is a Lie algebra isomorphism $\mathfrak g/\mathfrak n\cong \operatorname{Lie}(G/N)$. This proves both assertions. [/step]