By Lemma \[L:spin(2n+1)\], we get $g\in Z_{G}(H\cap\Ad(g)H)H$ for elements $g\in G$.

We have $$G^{\sigma}\subset\langle G,\sigma\theta\rangle/\langle\theta(-1)\rangle=\Ad(\theta)(\langle G,
\sigma\rangle/\langle-1\rangle).$$ Since $\langle G,\sigma\rangle/\langle-1\rangle\cong\O(8)$, we get $$\langle G,\sigma\theta\rangle/\langle\theta(-1)\rangle\cong\O(8).$$ From this we get an inclusion $\Spin(7)\hookrightarrow\O(8)$, which is just the spinor module of $\Spin(7)$. Write $$G'=\langle G,
\sigma\theta\rangle/\langle\theta(-1)\rangle,$$ and consider the pair $H=G^{\sigma}\subset G'$. We still write $g_{\theta}=\cos\theta+\sin\theta e_{7}e_{8}$, and write $[g_{\theta}]$ for its projection in $G'$. Write $A'=\{[g_{\theta}]:\theta\in[0,2\pi)\}$.

\[P:spin7-2\] For $g\in G'$, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails if and only if $$g\in
H[g_{\frac{k\pi}{4}}]H$$ ($k=1,3,5,7$).

Consider elements $g\in G'^{0}\sigma\theta$ and $g\in G'^{0}$ separately.

Firstly let $g\in G'^{0}\sigma\theta$. By the proof of Proposition \[P:spin7-1\], $H\backslash G'^{0}
\sigma\theta/H$ is a single double coset. Thus, it suffices to consider $g=\sigma\theta$. In this case, $H\cap\Ad(g)(H)=G^{\theta}$, and $$g\in Z_{G}(H\cap\Ad(g)H)\subset Z_{G}(H\cap\Ad(g)H)H.$$

Secondly let $g\in G'^{0}$. By Lemma \[L:symmetric2\] one shows that $G'^{0}=HA'H$. Then, it suffices to consider $g=[g_{\theta}]$ ($\theta\in[0,2\pi)$). When $\theta=0$ or $\pi$, $$g=
[g_{\theta}]\in Z_{G'}\subset Z_{G'}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$ When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, we have $$g=[g_{\theta}]=[\theta(-1)g_{\theta}]=[\pm{e_1\dots e_{6}}]\in H
\subset Z_{G'}(H^{g^2})H.$$

Now assume $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. In general, for $x\in H\subset G$, $[x]\in\Ad(g)H$ if and only if $g_{\theta}^{-1}xg_{\theta}\in H\cup H\theta(-1)$. This is also equivalent to $$\sigma(g_{\theta}^{-1}xg_{\theta})=g_{\theta}^{-1}xg_{\theta}$$ or $$\sigma(g_{\theta}^{-1}x
g_{\theta}\theta(-1))=g_{\theta}^{-1}xg_{\theta}\theta(-1).$$ Due to $(\sigma\theta(-1))\theta(-1)=-1$, the above is equivalent to $$[g_{2\theta},x]=\pm{1}.$$ The last condition just means the 