[proofplan] We prove the contrapositive of the desired local uniqueness statement. If every neighbourhood of $0$ in the transverse subspace $\mathfrak m$ contains a nonzero element whose exponential lies in $H$, then we construct a sequence $Y_k\to 0$ in $\mathfrak m$ with $\exp(Y_k)\in H$. After normalising and passing to a convergent subsequence, integer powers of $\exp(Y_k)$ approximate every point on the one-parameter subgroup generated by the limiting direction. Closedness of $H$ then forces that limiting direction to lie in $\mathfrak h$, contradicting the direct-sum condition $\mathfrak h\cap\mathfrak m=\{0\}$. [/proofplan] [step:Choose a sequence of nonzero transverse elements whose exponentials lie in $H$] Choose an auxiliary Euclidean norm $|\cdot|$ on the finite-dimensional real [vector space](/page/Vector%20Space) $\mathfrak m$. Suppose, toward a contradiction, that no open neighbourhood $V_{\mathfrak m}\subset\mathfrak m$ of $0$ satisfies \begin{align*} H\cap \exp(V_{\mathfrak m})=\{I\}. \end{align*} For each $k\in\mathbb N$, define the open ball \begin{align*} B_{\mathfrak m}(0,1/k):=\{Y\in\mathfrak m:|Y|<1/k\}. \end{align*} By the negation of the desired conclusion, there exists \begin{align*} Y_k\in B_{\mathfrak m}(0,1/k) \end{align*} such that $Y_k\ne 0$ and $\exp(Y_k)\in H$. Hence $Y_k\to 0$ in $\mathfrak m$. [guided] We begin by turning the failure of the theorem into a sequence. The topology on the finite-dimensional vector space $\mathfrak m$ may be described by any Euclidean norm, so fix one and denote it by $|\cdot|$. If the theorem were false, then every neighbourhood of $0$ in $\mathfrak m$ would contain some element $Y$ with $\exp(Y)\in H$ that is not allowed by the desired equality. The element $Y=0$ is not useful because $\exp(0)=I\in H$ always. Therefore the failure of \begin{align*} H\cap \exp(V_{\mathfrak m})=\{I\} \end{align*} means that in every neighbourhood of $0$ there is a nonzero transverse element whose exponential belongs to $H$. For each $k\in\mathbb N$, define \begin{align*} B_{\mathfrak m}(0,1/k):=\{Y\in\mathfrak m:|Y|<1/k\}. \end{align*} This is an open neighbourhood of $0$ in $\mathfrak m$. By the assumed failure, we may choose \begin{align*} Y_k\in B_{\mathfrak m}(0,1/k) \end{align*} with $Y_k\ne 0$ and $\exp(Y_k)\in H$. Since $|Y_k|<1/k$, the sequence satisfies $Y_k\to 0$ in $\mathfrak m$. [/guided] [/step] [step:Pass to a unit limiting direction in the complement $\mathfrak m$] For each $k\in\mathbb N$, define the unit vector \begin{align*} u_k:=\frac{Y_k}{|Y_k|}\in\mathfrak m. \end{align*} Then $|u_k|=1$. The unit sphere $S_{\mathfrak m}:=\{Y\in\mathfrak m:|Y|=1\}$ is compact because $\mathfrak m$ is finite-dimensional. Passing to a subsequence and relabelling, there exists $Y\in S_{\mathfrak m}$ such that $u_k\to Y$. In particular $Y\in\mathfrak m$ and $|Y|=1$. [/step] [step:Approximate each real multiple of the limiting direction by integer multiples of $Y_k$] Fix $t\in\mathbb R$. For each $k\in\mathbb N$, set \begin{align*} r_k:=|Y_k|. \end{align*} Then $r_k>0$ and $r_k\to 0$. Choose an integer $n_k\in\mathbb Z$ satisfying $|n_k-t/r_k|\leq 1/2$. Multiplying by $r_k$ gives $|n_kr_k-t|\leq r_k/2$. Thus $n_kr_k\to t$. Since $Y_k=r_ku_k$ and $u_k\to Y$, we obtain $n_kY_k=(n_kr_k)u_k\to tY$ in $\mathfrak g$. [guided] The goal is to prove that $\exp(tY)\in H$ for every real number $t$. We only know that $\exp(Y_k)\in H$, so the available elements of $H$ are integer powers of $\exp(Y_k)$. This is why we choose integers $n_k$ so that the scalar multiples $n_kY_k$ converge to $tY$. Define \begin{align*} r_k:=|Y_k|. \end{align*} Since $Y_k\ne 0$, each $r_k$ is positive. Since $Y_k\to 0$, we have $r_k\to 0$. For the fixed real number $t$, choose an integer $n_k\in\mathbb Z$ with \begin{align*} |n_k-t/r_k|\leq 1/2. \end{align*} Such an integer exists because every real number lies within distance $1/2$ of some integer. Multiplying the displayed inequality by the positive number $r_k$ gives \begin{align*} |n_kr_k-t|\leq r_k/2. \end{align*} Since $r_k\to 0$, it follows that $n_kr_k\to t$. Now use the normalisation $u_k=Y_k/|Y_k|$, equivalently $Y_k=r_ku_k$. Then \begin{align*} n_kY_k=(n_kr_k)u_k. \end{align*} The first factor converges to $t$, and the second factor converges to $Y$. Scalar multiplication is continuous in the finite-dimensional vector space $\mathfrak g$, so \begin{align*} n_kY_k\to tY. \end{align*} This produces the desired infinitesimal approximation to the entire line $\mathbb RY$. [/guided] [/step] [step:Use subgroup closure to force the limiting one-parameter subgroup into $H$] Because $\exp(Y_k)\in H$ and $H$ is a subgroup, the integer power $\exp(Y_k)^{n_k}$ belongs to $H$ for every $k\in\mathbb N$, including when $n_k<0$. Since each $Y_k$ is a matrix and commutes with itself, the exponential law for one matrix, [citetheorem:8778], gives $\exp(Y_k)^{n_k}=\exp(n_kY_k)$. The matrix exponential is continuous, and $n_kY_k\to tY$, so $\exp(n_kY_k)\to \exp(tY)$ in $GL(n,\mathbb C)$. Since $Y_k,Y\in\mathfrak g$ and $n_k,t\in\mathbb R$, the defining property of the matrix [Lie algebra](/page/Lie%20Algebra) gives $n_kY_k,tY\in\mathfrak g$ and hence $\exp(n_kY_k),\exp(tY)\in G$. The sequence $\exp(n_kY_k)$ lies in $H$, and $H$ is closed in $G$, hence $\exp(tY)\in H$. Because $t\in\mathbb R$ was arbitrary, $\exp(tY)\in H$ for every $t\in\mathbb R$. [/step] [step:Contradict the direct-sum decomposition of $\mathfrak g$] By the definition of the infinitesimal subgroup, \begin{align*} \mathfrak h=\{X\in\mathfrak g:\exp(tX)\in H\text{ for every }t\in\mathbb R\}. \end{align*} The preceding step proves that $Y\in\mathfrak h$. The construction of $Y$ also gives $Y\in\mathfrak m$. Therefore $Y\in\mathfrak h\cap\mathfrak m$. Since $\mathfrak g=\mathfrak h\oplus\mathfrak m$, we have $\mathfrak h\cap\mathfrak m=\{0\}$, so $Y=0$. This contradicts $|Y|=1$. Hence the supposition was false, and there exists an open neighbourhood $V_{\mathfrak m}\subset\mathfrak m$ of $0$ such that \begin{align*} H\cap \exp(V_{\mathfrak m})=\{I\}. \end{align*} [/step]