$G$.

Prove by induction on $\dim G$. First, using Kac coordinate one shows that there are only finitely many conugacy classes of subgroups $H$ of $G$ of the form $H=Z_{G}(g)$ for some $g\in G$.

If $A\subset Z_{G}(G^{0})$, then $Z_{G}(A)\supset G^{0}$. As $G$ is a compact Lie group, $G/G^{0}$ is a finite group. Thus, there are only finitely many subgroups containing $G^{0}$.

If $A\not\subset Z_{G}(G^{0})$, take an $g\in A-Z_{G}(G^{0})$ and set $G'=Z_{G}(g)$. Then, $A\subset Z_{G}(A)\subset
G'$. Thus, $Z_{G}(A)=Z_{G'}(A)$. In the beginning we have shown that the conjugacy class of $G'=Z_{G}(g)$ has only finitely many possibilities. Since $g\not\in Z_{G}(G^{0})$, we have $G'\not\supset G^{0}$. Thus, $\dim G'<\dim G$. By induction subgroups $H$ of $G'$ of the form $H=Z_{G'}(A)$ and $A$ abelian have only finitely many possibilities. This finishes the proof.

\[P:acceptable2\] If a compact Lie group $G$ is acceptable, then for any quasi-torus $H$ and element-conjugate homomorphisms $\phi_{1},\phi_{2}:H\rightarrow G$, $\phi_1$ and $\phi_2$ are globally conjugate.

As $\phi_1$ and $\phi_2$ are element-conjugate, we have $\ker\phi_1=\ker\phi_2$. Considering $H/\ker\phi_1$ instead, we may assume that $\phi_1$ and $\phi_2$ are injections. Write $$H^{0}[n]=\{x\in H^{0}:x^{n}=1\}$$ for the group of $n$-torsion elements in $H^{0}$. For any $m\geq 1$, set $$H_{m}=Z_{G}(\phi_{1}(H^{0}[m])).$$ Then, $$H_1\supset H_2
\supset\cdots H_{m}\supset\cdots.$$ By Lemma \[L:centralizer1\], there exists 