Every abelian normal subgroup of a connected and compact Lie group lies in the center 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.
 A: To give a more elementary solution: Let $H^0$ be the identity component of $H$.
$H^0$ is compact, connected and abelian, hence it is a torus and is contained in a maximal torus $T$. Thus,
$$H^0 = g H^0 g^{-1} \subset gTg^{-1}$$
Since all maximal tori are conjugates of each other, $H^0$ is contained in all of them and hence in the center of $G$ (This is Theorem 2.3 part (iii) pg. 165 in Theodor Brocker's Representation of Compact Lie Groups)
In the case that $H$ is disconnected, let $h\in H\backslash H^0$ and let $T_h$ be a maximal torus that contains $h$. Observe that $H/H^0$ is a discrete normal subgroup of $G/H^0$ and hence it is abelian. From this we get that for all $g\in G$ there exists an $h_0 \in H^0$ such that:
$$g h g^{-1} = h\, h_0$$
Finally, this means that the set $c_G(h) = \{ g h g^{-1}: \ \forall g\in G\}$ is contained in $T_h$ because both $h$ and $h_0$ are contained in $T_h$. With the same argument as before,
$$c_G(h) = g\, c_G(h) \, g^{-1} \subset gT_h g^{-1}$$
and this means that $c_G(h)$ is contained in all maximal tori of $G$.
A: 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
