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For a closed subgroup $H$ of a Lie group, this set is a real Lie subalgebra. Scalar closure is immediate; the Lie product formula in a Lie group and closedness of $H$ give addition. It is a closed vector subspace, also directly by continuity of the Exponential map of a Lie group. Conjugating its one-parameter subgroups by elements of $H$ preserves it. Differentiating the Adjoint representation of a Lie group along $\exp (sX)$ gives $[X,Y]\in \mathfrak{h}$ for $X,Y\in \mathfrak{h}$. After the closed-subgroup theorem establishes smoothness, $\mathfrak{h}$ is exactly the tangent space $T_{e}H$.

DefineIt contains zero, and $X\in \mathfrak{h}$ implies $aX\in \mathfrak{h}$ for every real scalar $a$, by replacing $t$ with $at$.

For $X,Y\in \mathfrak{h}$, we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let $Z(s)=\log (\exp (sX)\exp (sY))$. Smooth multiplication has differential $(U,V)\mapsto U+V$ at the identity, so Taylor's theorem givesFor each fixed real $t$ and all sufficiently large positive integers $m$,Every term belongs to $H$, and $H$ is closed, proving $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a real vector subspace. It is also closed: if $X_j\to X$ with $X_j\in \mathfrak{h}$, continuity of the Exponential map of a Lie group gives $\exp (tX)\in H$ for each $t$.

For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently ${\mathrm{Ad}}_{\exp (sX)}Y\in \mathfrak{h}$. For $s\ne 0$, linearity puts $({\mathrm{Ad}}_{\exp (sX)}Y-Y)/s$ in $\mathfrak{h}$. Its limit as $s\to 0$ is ${\mathrm{ad}}_{X}Y=[X,Y]$. Closedness of $\mathfrak{h}$ therefore gives $[X,Y]\in \mathfrak{h}$, proving that the Lie algebra of a closed subgroup is a Lie subalgebra of $\mathfrak{g}$.