# On closed abelian reductive subgroups of Real reductive groups

Hello everybody. I would first like to apologise for the basic question; I'm not expert on Lie Theory. Can someone please help me with the following questions

Let $\mathrm{G}=\mathrm{K} \exp(\mathfrak{p})$ be a Real reductive group (as in Knapp's Lie Groups: Beyond an Introduction) and let $\mathfrak{a}$ be a maximal Lie subalgebra of $\mathfrak{g}=\operatorname{Lie}(\mathrm{G})$ such that $\mathfrak{a} \subseteq \mathfrak{p}$. Since ${[\mathfrak{p},\mathfrak{p}]} \subset \mathfrak{k}$ ($\mathfrak{k}=\operatorname{Lie}(\mathrm{K})$), the Lie algebra $\mathfrak{a}$ is commutative. Let $A:=\exp(\mathfrak{a})$ denote the corresponding commutative Lie subgroup of $\mathrm{G}$.

My questions are:

1. Is $\mathrm{A}$ a closed subgroup of $\mathrm{G}$?
2. Is $\mathrm{A}$ a reductive subgroup of $\mathrm{G}$?
3. In otherwise, Can someone give some examples to show that 1. and 2. are not always true?

Any answer or comment will be greatly appreciated. Thanks.

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Answers are: 1. yes, 2. yes. A is a maximal split torus (or rather the connected component of its real points). – doug Nov 29 '11 at 20:11