Let $G=SO(3)$ and $\theta:G\to G$ defined by $\theta(g)=JgJ$ where $J=\text{diag}(-1,1,1)$. Prove that, $\theta(g)\in Kg^{-1}K$ where $K=SO(2)$ Here $K=\left\{\begin{pmatrix}1 & 0\\ 0 & T_1\end{pmatrix}:\ T_1\in SO(2)\right\}$. $K$ is a compact subgroup of $G$. The above problem is part of the proof of $(G,K) $ is Gelfand pair.
I've already prove that the above $\theta$ is involutive automorphism of $G$. But I'm unable to show that $\theta(g)\in Kg^{-1}K\ \forall g\in G$.
I know that $G=KAK$ where $A=\left\{\begin{pmatrix}\cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta\end{pmatrix}:\ \theta\in[0,2\pi)\right\}$. Can I anyone use this to attempt the problem?
Can anyone suggest me a wayout? Thanks for help in advance.
 A: Given $g \in G = SO(3)$, I'll use the notation $g_0$ to refer to the top right entry of $g$.  I'll also use transpose notation $(a,b,c)^t = \begin{bmatrix} a\\b\\c\end{bmatrix}$ to avoid writing such large lines.
Note that $g^{-1} = g^t$ and $(g^t)_0 = g_0$.  Moreover, $\theta(g)_0 = g_0$  Thus, what you want to show  follows from the following more general proposition.
Proposition:  Suppose $g,h\in G$ with $g_0  = h_0$.  Then there are $k_1,k_2\in K$ with $g = k_1 h k_2$.
Proof outline: $1$.  Given an vector $(a,b)^t$, there is a rotation matrix $R = \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$ with $R(a,b)^t =  (\lambda,0)^t$ for $\lambda = \sqrt{a^2 + b^2}\geq 0$.  (When $\lambda > 0$, $\theta$ is uniquely determined).
$2$.  Given $g\in G$, there is an element $k_1\in K$ with the property that the first column of $k_1g$ is $(g_0, \lambda ,0)^t$ with $\lambda = \sqrt{1-|g_0|} \geq 0$.
$3$.  Given $g\in G$ having the form of 2. there is a $k_2\in G$ with $gk_2$ having first row $(g_0, \lambda, 0)$.  Moreover, note that $gk_2$ has the same first column as $g$.
Combining $2$ and $3$, we can find $k_1,k_2$ for which $k_1 g k_2$ has the form  $\begin{bmatrix} g_0 & \lambda & 0\\ \lambda & * & *\\ 0 & * & *\end{bmatrix}$.
$4a$.  Using orthgonality and $\det = 1$, show that if $g_0 \notin\{\pm 1\}$, then all four $\ast$ are determined by the other entries.
$4b$.  Using orthgonality and $\det  =1$, show that if $g_0\in \{\pm 1\}$, then we may further repick $k_1$ and $k_2$ so that $k_1 g k_2 = \operatorname{diag}(g_0,1,g_0)$.
$5$.  Thus, in either case, given $g\in G$, we can find $k_1,k_2$ with $k_1 g k_2$ in some "normal form" which is completely determined by $g_0$.  Thus, if $h\in G$ with $h_0 = g_0$, then $g$ and $h$ have the same normal form.  Conclude.  $\square$
