$\O(2n+1)\subset\SO(2n+2)$, the pair $\SO(2n+1)\subset\SO(2n+2)$ is an “SCF” pair.

\[P:O(2n+1)-2\] For any $n\geq 1$, $\SO(2n+1)$ is and “SCF” subgroup of $\SO(2n+2)$.

Following the notation as in the proof of Proposition \[P:O(2n+1)-1\]. We have $H=\O(2n+1)$, and $H^{0}=\SO(2n+1)$.

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, we have $$H\cap\Ad(g_{\theta})H
=H^{g_{\theta}^2}=S(\O(2n)\!\times\!\O(1)\!\times\!\O(1)).$$ Then, $$H^{0}\cap\Ad(g_{\theta})H=
S(\O(2n)\!\times\!\O(1)).$$ By this, $$H^{0}\cap\Ad(g_{\theta})H^0=\SO(2n)\textrm{ or }
S(\O(2n)\!\times\!\O(1)).$$ By calculation one sees that $\diag\{I_{2n-1},-1,-1,1\}\not\in
\Ad(g_{\theta})H^0$. Thus, $$H^{0}\cap\Ad(g_{\theta})H^0=\SO(2n).$$ Hence, $$g_{\theta}\in
Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, by direct calculation one sees that $H^{0}
\cap\Ad(g_{\theta})H^0=\SO(2n)$. Thus, $$g_{\theta}\in Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)
\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

When $\theta=0$ or $\pi$, $$g_{\theta}\in Z_{G}H^0\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

[*The pair $\Spin(2n+1)\subset\Spin(2n+2)$.*]{} Let $n\geq 1$, $G=\Spin(2n+2)$ and $\theta
=\Ad(e_{2n+2})$. Then, $\theta^2=1$. Write $H=G^{\theta}$. Then, $H=\Spin(2n+1)$. Write $$g_{\theta}
=\cos\theta+\sin\theta e_{2n+1}e_{2n+2}$$ ($\theta\in\mathbb{R}$). By Lemma \[L:symmetric2\], we have $G=HAH$ for $$A=\{g_{\theta}:\theta\in[0,2\pi)\}.$$

\[L:spin(2n+1)\] For any $n\geq 1$, $\Spin(2n+1)$ is and “SCF” subgroup of $\Spin(2n+2)$.

By Lemma \[L:symmetric4\], it suffices to show that $g\in Z_{G}(H^{g^2})H$ for any $g=
g_{\theta}$ ($\theta\in[0,2\pi)$).

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, we have $g^2=g_{2\theta}$ and $H^{g^{2}}=\Spin(2n)$. Thus, $$g\in Z_{G}(H^{g^2})\subset Z_{G}(H^{g^2})H.$$

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, we have $e_1\dots e_{2n+2}\in Z_{G}\subset
Z_{G}(H^{g^2})$ and $(e_1\dots e_{2n+2})^{-1}g=\pm{e_1\dots e_{2n}}\in H$. Thus, $$g=(e_1\dots e_{2n+2})((e_1\dots e_{2n+2})^{-1}g)\in Z_{G}(H^{g^2})H.$$

When $\theta=0$ or $\pi$, $$g=g_{\theta}\in Z_{G}\subset Z_{G}(H^{g^2})H.$$

[*Pinned automorphisms.*]{}

\[T:pinned\] Let $G$ be a connected and simply-connected compact simple Lie group and $\theta$ be a (nontrivial) 