give a new proof for Theorem \[T:Griess\]. Our method is to to study the intersections $H\cap\Ad(g)H$ ($g\in G$ and the double coset decomposition $H\backslash G/H$. This consideration has independent interest.

\[L:SCF2\] Let $G$ be a compact Lie group and $H\subset G$ be a closed subgroup. Then $H$ is an “SCF" subgroup of $G$ if and only if $$\label{Eq2}g\in Z_{G}(H\cap\Ad(g)H)H$$ for any $g\in G$.

Necessarity. Suppose $H$ is an “SCF" subgroup of $G$. For any $g\in G$, set $$\Gamma=H\cap\Ad(g)H.$$ Let $\phi_1:\Gamma\rightarrow H$ be the inclusion map, and $\phi_2: \Gamma\rightarrow H$ be $\phi_2=\Ad(g^{-1})
\circ\phi_{1}$. Then, $\phi_1,\phi_2:\Gamma\rightarrow H$ and $\phi_2=\Ad(g^{-1})\circ\phi_1$. As $H$ is an “SCF" subgroup of $G$, there exists $h\in H$ such that $\phi_2=\Ad(h^{-1})\circ\phi_{1}$. Then, $\phi_1=\Ad(gh^{-1})\circ\phi_1$. Thus, $gh^{-1}\in Z_{G}(\Gamma)$, and $$g=(gh^{-1})h\in Z_{G}(H\cap\Ad(g)H)H.$$

Sufficiency. Suppose $g\in Z_{G}(H\cap\Ad(g)H)H$ for any $g\in G$. Let $\Gamma$ be a compact group and $\phi_1,\phi_2:\Gamma\rightarrow H$ be two homomorphisms with $\phi_1=\Ad(g)\circ\phi_2$ for some $g\in G$. Then, $\Im\phi_1\subset H$ and $$\Im\phi_1=\Im(\Ad(g)\circ\phi_2)\subset\Ad(g)H.$$ Hence, $\Im\phi_1\subset
H\cap\Ad(g)H$. By the assumption, we then have $$g\in Z_{G}(H\cap\Ad(g)H)H\subset Z_{G}(\Im\phi_1)H.$$ Thus, $g=g'h$ for some $g'\in Z_{G}(\Im\phi_1)$ and $h\in H$. Hence, $$\phi_2= \Ad(g^{-1})\circ\phi_1=
\Ad(h^{-1})\circ\Ad(g'^{-1})\circ\phi_1=\Ad(h^{-1})\circ\phi_1.$$ Therefore, $H$ is an “SCF" subgroup of $G$.

\[R:SCF2\] It is easy to see that if the condition (\[Eq2\]) in Lemma \[L:SCF2\] holds for 