KZ=K(H\cap Z)=K\cong\SU(3).$$ This shows $Z_{H}(g)=\Stab_{H}(X)$ by Lemma \[L:G2\]. If $g^2=1$, then $G^{g}=\overline{G}_{X}\cong\O(6)$ and $$Z_{H}(g)=H\cap G^{g}=H\cap\overline{G}_{X}.$$ By Lemma \[L:G2\] again, we get $Z_{H}(g)^{0}\cong\SU(3)$ and the index $$[Z_{H}(g),Z_{H}(g)^{0}]=2.$$

\[L:G2-2\] For any $g\in\exp(V)$, $H\cap\Ad(g)H=H^{g^{3}}$.

Consider $H=\G_2$ as a subgroup of $G'=\PSO(8)$. There is an order three element $\theta\in\Aut(G')$ such that $H=G^{\theta}$. Analogous to the proof of Lemma \[L:symmetric3\], one can show that $$H\cap\Ad(g)H=Z_{H}(\theta(g)^{-1}g).$$

Write $\mathfrak{q}_0$ for the orthogonal complement of ${\mathfrak{h}}_0$ in ${\mathfrak{g}}'_0$. Then, $\theta$ acts on $\mathfrak{q}_0$ as a linear transformation with $\theta^2+\theta+1=0$. For any $X\in V$, we show that $[\theta(X),X]=0$. Write $Y=[\theta(X),X]$. Write $\sigma\in\Aut(G')$ for an element with $G'^{\sigma}=G$. Then, $\sigma^2=1$ and $\sigma\theta\sigma^{-1}=\theta^{-1}$. Then, $$\theta(Y)=[\theta^{2}(X),\theta(X)]
=[-\theta(X)-X,\theta(X)]=[\theta(X),X]=Y$$ and $$\!\sigma(Y)\!=\![\sigma\theta(X),\!\sigma(X)]\!=\!
[\theta^{-1}\!\sigma(X),\!X]\!=\![\theta^{2}\!(X),\!X]\!=\![-\theta(X)\!-\!X,\!X]\!=\!-Y.$$ Since ${\mathfrak{g}}^{\theta}={\mathfrak{h}}\subset{\mathfrak{g}}^{\sigma}={\mathfrak{g}}$, we get $Y=\sigma(Y)=-Y$. Thus, $Y=0$.

Write $g=\exp(X)$ for some $X\in{\mathfrak{p}}_0$. Write $X'=X-\theta(X)$. Then, $\theta(X')=\theta(X)-\theta^{2}(X)
=X+2\theta(X)$. Thus, $2X'+\theta(X')=3X$. By this, $$Z_{H}(\theta(g)^{-1}g)=Z_{H}(\exp(X'))\subset
Z_{H}(\exp(3X))=Z_{H}(g^{3}).$$ Thus, $Z_{H}(g)\subset Z_{H}(\theta(g)^{-1}g)\subset Z_{H}(g^{3})$.

When $g^6\neq 1$, by Lemma \[L:G2-1\], we get $Z_{H}(g)=Z_{H}(g^3)\cong\SU(3)$. Thus, $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})$. When $g^2=1$, then $g=g^3$ and $Z_{H}(g)=Z_{H}(g^3)$. Thus, $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})$. When $g^3=1$, we have $g\in Z\cap K\subset H$. Thus, $\theta(g)
=g$ and $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})=H$. When $o(g)=6$, we have $g^2\in H$, $\theta(g)^{-1}g
=\theta(g^3)^{-1}g^{3}$, and $$Z_{H}(\theta(g)^{-1}g)=Z_{H}(\theta(g^3)^{-1}g^3)=Z_{H}(g^3).$$

It suffices to show $(G,H)=(\SO(7),\G_2)$ is an “SCF” pair. By Lemma \[L:SCF2\] and Lemma \[L:symmetric1\], we just need to show: for any $g\in\exp(V)$, 