Cartan's closed-subgroup theorem
/ kar-TAHN /
When is a subset of a Lie group automatically a Lie group in its own right, with no smoothness imposed by hand? The answer is startlingly clean: just being topologically closed is enough. You do not need to check that the subgroup is a manifold or that the operations are smooth — closedness alone forces a unique smooth structure into existence. This is one of the most useful labor-saving theorems in the subject.
Precisely, Cartan's closed-subgroup theorem (also called the Cartan-von Neumann theorem) states: if H is a subgroup of a Lie group G that is closed as a subset (topologically), then H is an embedded Lie subgroup — a submanifold of G that is itself a Lie group with the subspace topology, and the inclusion is a smooth embedding. The proof shows the Lie algebra of H is the set h = {X in g : exp(tX) in H for all t in R}, which turns out to be a linear subspace and a subalgebra, and that exp maps a neighborhood of 0 in h diffeomorphically onto a neighborhood of e in H, supplying the chart. This is exactly why the classical matrix groups (O(n), SL(n), U(n), Sp(n), ...) are Lie groups: each is a CLOSED subgroup of some GL(n), cut out by continuous equations like A^T A = I or det A = 1, so the theorem hands you the manifold structure for free.
The theorem is the standard tool for recognizing Lie subgroups, building homogeneous spaces G/H (which need H closed to be a manifold), and verifying that stabilizers and kernels are Lie subgroups. The crucial honesty point is the word CLOSED: it cannot be dropped. The irrational winding line on the torus is a subgroup of T^2 and is an immersed Lie subgroup, but it is dense, not closed, and not embedded — so non-closed subgroups can fail to be embedded submanifolds. Closedness is precisely the hypothesis that upgrades an abstract algebraic subgroup to a genuine smooth submanifold.
O(n) is the zero set of the map A |-> A^T A - I, a continuous (indeed polynomial) condition, so it is a closed subset of GL(n, R). Cartan's theorem instantly upgrades it to an embedded Lie subgroup of dimension n(n-1)/2 — no separate verification that it is a smooth manifold is needed.
Closed in GL(n) by the equation A^T A = I, so O(n) is a Lie group for free.
Closedness is essential and cannot be weakened to 'subgroup'. The dense irrational line in the 2-torus is a subgroup that is NOT closed and is only immersed, not embedded.