$(\U(n),\O(n))$, $(\U(2n),\Sp(n))$, $(\E_6,\F_4)$. For any $g\in Z_{G}$, we have $$g\in Z_{G}(H^{g^2})\subset Z_{G}(H^{g^{2}})H.$$ For any $g\in A-Z_{G}$, by Lemma \[L:symmetric5\] there exists $X\in{\mathfrak{a}}_0$ such that $$g^{2}=\exp(2X)\textrm{ and }G^{g^{2}}=
\Stab_{G}(X).$$ Write $z=\exp(X)$ and $k=\exp(-X)g$. By $G^{g^{2}}=\Stab_{G}(X)$, we get $z\in Z_{G}
(H^{g^{2}})$. By $g^{2}=\exp(2X)$, we get $k^2=1$. Then, $\theta(k)=k^{-1}=k$. Thus, $k\in H$. Therefore, $g=zk\in Z_{G}(H^{g^{2}})H$. By Lemma \[L:SCF2\], the conclusion follows.

\[P:SO(2n+1)\] For any $n\geq 1$, $\SO(2n+1)$ is an “SCF" subgroup of $\U(2n+1)$.

Due to $\O(2n+1)=\SO(2n+1)\times\{\pm{I}\}$, $\SO(2n+1)$ is an “SCF” subgroup of $\O(2n+1)$. Then, the conclusion follows from Theorem \[T:Griess\](1).

$\G_2$
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Let $(G,H)=(\SO(7),\G_2)$. Write $V={\mathfrak{p}}_0$. It is clear that $V$ is an irreducible real representation of $\G_2$ with $V\otimes_{\mathbb{R}}\mathbb{C}$ a nontrivial irreducible representation of $\G_2$ with minimal dimension.

For any $X\in V$, write $G_{X}=\Stab_{\SO(7)}(X)$ and $$\overline{G}_{X}=\{g\in\SO(7), g\cdot X=\pm{X}.\}$$

\[L:G2\] For any $0\neq X\in V$, we have $HG_{X}=G$, $\Stab_{H}(X)=H\cap G_{X}\cong\SU(3)$, and $\Stab_{H}(X)$ is an index $2$ subgroup of $H\cap\overline{G}_{X}$.

Write $M$ for the unit sphere in $V$. Then, $M\cong S^{6}$, and $G$ acts on $M$ transitively. Note that a proper closed subgroup of $H$ has dimension at most $8$. Thus, $\dim H\cdot X\geq 14-8=6=\dim M$ for any $X\in M$. By the connectedness of $M$. We see that $H$ acts on $M$ transitively. Then, 