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Every closed subgroup of a finite-dimensional Lie group is an embedded submanifold and a Lie subgroup. Closedness alone forces the compatible smooth structure; smoothness need not be assumed in advance. The proof constructs the Lie algebra of a closed subgroup from one-parameter subgroups, then uses local product coordinates for a closed subgroup.

Choose a real vector subspace $\mathfrak{m}$ complementary to $\mathfrak{h}$, so $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$. The smooth maphas differential $(U,V)\mapsto U+V$ at $(0,0)$, an isomorphism. The inverse function theorem makes $F$ a diffeomorphism from a sufficiently small product neighborhood onto an open neighborhood of the identity in $G$.

There is a neighborhood of zero in $\mathfrak{m}$ in which $\exp U\in H$ implies $U=0$. Otherwise choose nonzero $U_j\in \mathfrak{m}$ tending to zero with $\exp U_j\in H$. A normalized subsequence tends to a unit vector $U\in \mathfrak{m}$. The limit directions of a closed subgroup from part (a) give $U\in \mathfrak{h}$, contradicting $\mathfrak{m}\cap \mathfrak{h}=0$. Shrink the product neighborhood so that this exclusion holds throughout its first factor.

If $F(U,V)\in H$, then $\exp V\in H$ by the definition of $\mathfrak{h}$, and hence $\exp U=F(U,V)\exp (-V)\in H$. The exclusion forces $U=0$. Conversely, every $F(0,V)=\exp V$ belongs to $H$. Thus the local product coordinates for a closed subgroup identify $H$ near the identity exactly with the smooth coordinate slice $U=0$.

Left translation by each element of $H$ gives the same description around that element, so $H$ is an embedded submanifold of $G$ in its subspace topology. The multiplication and inversion maps of $G$ restrict to smooth maps on this embedded submanifold, making $H$ a Lie group and an embedded Lie subgroup. Its tangent space at the identity is $\mathfrak{h}$, because $d{\exp }_0$ is the identity on this coordinate slice. It remains closed by hypothesis. This proves the closed-subgroup theorem, with no prior smoothness assumption on $H$.