= Solution
A real <Lie group> is a finite-dimensional real <smooth manifold> with a <group> structure whose multiplication and inversion are smooth. Its <tangent space> at the <identity element> $e$ is the vector space $\mathfrak g=T_eG$ of velocities of smooth curves through $e$. For the <general linear group> $\mathrm{GL}_n(\mathbb R)$ it identifies with $M_n(\mathbb R)$. The <matrix exponential> and local <matrix logarithm> are
$$
\exp X=\sum_{r=0}^\infty\frac{X^r}{r!},\qquad\log(I+A)=\sum_{r=1}^\infty\frac{(-1)^{r+1}A^r}{r}\quad(\|A\|<1).
$$
The exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp(-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.
For a general $G$, the assumed <Exponential map of a Lie group> with $d\exp_0=I$ has a smooth local inverse by the <inverse function theorem>, giving a <local exponential chart>. Define the <Lie bracket from local group commutators> by
$$
\boxed{[X,Y]=\left.\frac{\partial^2}{\partial s\,\partial t}\right|_{(0,0)}\log\!\left(\exp(sX)\exp(tY)\exp(-sX)\exp(-tY)\right).}
$$
For small $s,t$ the <group commutator> lies in that chart. More explicitly, the map $F(u,v)=\log(\exp u\exp v\exp(-u)\exp(-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a <bilinear map> of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log(g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar <commutator> $XY-YX$.
For $g\in G$, let $c_g(h)=ghg^{-1}$ and define the <Adjoint representation of a Lie group> by $\operatorname{Ad}_g=d(c_g)_e\in\mathrm{GL}(\mathfrak g)$. It is smooth and satisfies $\operatorname{Ad}_{gh}=\operatorname{Ad}_g\operatorname{Ad}_h$. Its <derived representation> is $\operatorname{ad}=d(\operatorname{Ad})_e:\mathfrak g\to\operatorname{End}(\mathfrak g)$. Naturality of the <Exponential map of a Lie group> under conjugation gives
$$
\exp(sX)\exp(tY)\exp(-sX)=\exp\!\left(t\operatorname{Ad}_{\exp(sX)}Y\right).
$$
At $t=0$, the derivative of the logarithm of the commutator is $\operatorname{Ad}_{\exp(sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d\log_e=I$. Differentiate in $s$ to conclude
$$
\boxed{\operatorname{ad}(X)Y=[X,Y].}
$$
To prove the <Jacobi identity> without assuming it in the bracket construction, first establish that the <differential of a Lie group homomorphism preserves Lie brackets>. For a homomorphism $f:G\to H$, the allowed exponential identity gives $\log_H(f(\exp_Gu))=df_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $df_e([X,Y])=[df_eX,df_eY]$. In particular $f=\operatorname{Ad}:G\to\mathrm{GL}(\mathfrak g)$ gives
$$
\operatorname{ad}[X,Y]=[\operatorname{ad}X,\operatorname{ad}Y].
$$
Applying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry proves
$$
\boxed{[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0.}
$$
Neither injectivity of $\operatorname{ad}$ nor the existence of a global logarithm was used.
Finally let $H$ be a normal <Lie subgroup> of $G$, with its corresponding <Lie algebra> $\mathfrak h\subseteq\mathfrak g$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak h$: $\operatorname{Ad}_g\mathfrak h=\mathfrak h$. Differentiate along $g=\exp(tX)$ to get $[X,Y]=\operatorname{ad}(X)Y\in\mathfrak h$ for every $X\in\mathfrak g$ and $Y\in\mathfrak h$. Thus \b[the <Lie algebra of a normal Lie subgroup> is an <ideal of a Lie algebra>]. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.
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