Let $G$ be a compact non-connected nilpotent Lie subgroup of $O(n)$. We know that $G\_0$, its identity component, is always a torus. Is it true that $G\_0$ is always central in $G$?


What about general $G$ (which may not be subgroup of $O(n)$)?



Compact non-connected nilpotent Lie subgroup of $O(n)$?
If this is your question, yes it's true that for every nilpotent compact Lie group $G$, $G\_0$ is central. 


Indeed as you already noticed, $G\_0$ is abelian, so the action by conjugation on $G\_0$ factors through the finite group $F=G/G\_0$. Let $V$ be the universal covering of $G\_0$: this is a vector group (= finite-dimensional real vector space), and the action of $F$ lifts to $V$ (as $V$ can be identified to the Lie algebra of $G$). Then $V$ splits as $V^F\oplus W$, with $V^F$ the vectors fixed by $F$, and $W$ is its unique $F$-invariant complement. Then $[F,W]=W$. It follows that, if $M$ is the closure of the image of $W$ in $G\_0$, we have $[G,M]=M$. Since $G$ is nilpotent, this forces $M=1$, hence $W=0$, which means that $V=V^F$ and thus $G\_0$ is central.


