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Let be $G$ a Lie groupe and $H$ a Lie subgroup,if $H$ is abelian then the closure $\bar{H}$ of $H$ is also abelian.
Please give a proof of this proposition.

Antema Kely
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1 Answers1

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This is simply because multiplication is continuous. Take $ a_k \to a, b_k \to b$ where $\{a_k\}_{k \in \mathbb N}, \{b_k\}_{k \in \mathbb N}$ are sequences of elements of $H$.

Then $ab = \lim\limits_{k \in \mathbb N} a_k \cdot b_k = \lim\limits_{k \in \mathbb N} b_k \cdot a_k = ba$.