Show that every abelian normal subgroup $H$ of a connected and compact Lie group $G$ lies in the center of $G$.
It may be helpfull that if $f:G\to H$ is a surjective lie group homomorphism with $H$ a abelian group, then the restriction $f|_T$ of $f$ to any maximal torus $T$ is also surjective.
But here is what I am trying: take $h\in H$ and $g\in G$. As $H$ is normal, one has that there is a $h'\in H$ s.t. $$gh=h'g$$ My attempt is to show that $h'=h$.
Let $T$ be a maximal torus in $G$. One has that there is $r\in G\setminus T$ and $t\in T$ s.t. $$rtr^{-1}=h.$$ As $H$ is normal, $t\in H\cap T$.
Thus, $$(gr)t(gr)^{-1}=h'.$$
But I'm stuck here. Also I'm not using that $H$ is abelian.
It results from the structure of connected compact Lie group, since $G$ is the quotient $S\times T$ of a semi simple group $S$ and a torus $H$ by a central subgroup. If $p$ is the quotient map, $p^{-1}(H)$ is normal and abelian so contained in $e\times T$.
https://en.m.wikipedia.org/wiki/Compact_group