Is subgroup of a Lie group (in algebraic sense) automatically a manifold?

We all know an open subset of a manifold $M$ is also a manifold (but may not a submanifold sinse we havn't ask the inclusion map be an immersion or even embdedding). Since a Lie group $G$ is also a smooth manifold, so analogically, it is natural to ask whether its subgroup (in algebraic sense) also a manifold ?

Here let me first talk about my understanding :

Since we have Caratan's theorem, so if the subgroup $H$ of Lie group $G$ admits a manifold structure, then it satisfies the dinition of Lie group if and only if $H $ is a closed groupn since $H\times H \rightarrow H$ makes sense and it is smooth if we restricts the smooth map on $G_{|H}$.

Also I think it may lead some confusion by meaning "analogy" because open subset is a topological notion but not a group notion. So it seems to be strange to ask Lie subgroup meanwhile has a manifold structure which is a topological sense.

I'm not sure of my understanding. So any good example is welcome. Thanks for your suggestions !


1 Answer 1


Take the Lie group $(\mathbb{R},+)$, one may check that the subgroup $(\mathbb{Q},+)$ is not a manifold (with the subspace topology). Quotienting everything by $\mathbb{Z}$ gives a compact example.

  • $\begingroup$ Clear answer,thanks! $\endgroup$
    – LSY
    Mar 3, 2020 at 8:40
  • $\begingroup$ By the way, once a subgroup is also a manifold, then by Cartan's theorem,it is closed, so any non-closed subgroup also a case. Is it correct? $\endgroup$
    – LSY
    Mar 3, 2020 at 8:42
  • 1
    $\begingroup$ Yes, in fact the example I gave is non-closed. $\endgroup$
    – Nick L
    Mar 3, 2020 at 8:49

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