7.29], we have $${\mathfrak{p}}_0=\Ad(H^{0}){\mathfrak{a}}_0=\{\Ad(k)X:k\in H^{0},X\in{\mathfrak{a}}_0\}.$$ Then, by Lemma \[L:symmetric1\] we get $$G\!=\!H^{0}\exp({\mathfrak{p}}_0)\!=\!H^{0}\exp(\Ad(H^{0}){\mathfrak{a}}_0)\!=\!
H^{0}\exp({\mathfrak{a}}_0)H^{0}\!=\!H^{0}AH^{0}.$$

\[L:symmetric3\] For any $g\in\exp({\mathfrak{p}}_0)$, $H\cap\Ad(g)H=H^{g^{2}}$.

By definition $H=G^{\theta}$. Thus, $x\in H\cap\Ad(g)H$ if and only if $\theta(x)=x$ and $\theta(g)^{-1}
\theta(x)\theta(g)=\theta(g^{-1}xg)=g^{-1}xg$. This is also equivalent to: $x\in H$ and $x\in G^{\theta(g)g^{-1}}$. Due to $g\in\exp({\mathfrak{p}}_0)$, we have $\theta(g)=g^{-1}$. Therefore, we get $H\cap\Ad(g)H=H^{g^{2}}$.

Combining Lemma \[L:SCF2\], Lemma \[L:symmetric2\] and Lemma \[L:symmetric3\], we get the following statement.

\[L:symmetric4\] Assume that $\theta$ is an involutive automorphism of $G$, and $H=G^{\theta}$. Then, $H$ is an “SCF" subgroup of $G$ if and only if $g\in Z_{G}(H^{g^{2}})H$ for any $g\in A$.

\[L:symmetric5\] Let $(G,H)$ be one of the following pairs: $$(\U(n),\!\O(n)),\ (\U(2n),\!\Sp(n)),\ (\E_6,\!\F_4).$$ For any $g\in A-Z_{G}$, there exists $X\in{\mathfrak{a}}_0$ such that $g^{2}=\exp(2X)$ and $G^{g^{2}}=\Stab_{G}(X)$.

We know that the set of nonzero $A$ weights appearing in ${\mathfrak{g}}$ forms a restricted root system (cf. [@Knapp Page 370]). The centralizer $G^{g}=Z_{G}(g)$ ($g\in A$) and the stabilizer $\Stab_{G}(X)$ ($X\in{\mathfrak{a}}_0$) could be calculated from the evaluation of restricted roots on $g$ (or on $X$). The restricted root system for the pairs $(\U(n),\O(n))$, $(\U(2n),\Sp(n))$, $(\E_6,\F_4)$ are of types $\A_{n-1}$, $\A_{n-1}$, $\A_{2}$ respectively. One could show the conclusion by a case by case verification.

Let $(G,H)$ be one of the following pairs: 