Show that $G_E=\mathbb{S}^1\times \mathbb{S}^1$ I am stuck at the problem below.
(I am using two identification $\mathbb{R}^4=\mathbb{H}$ and $\mathbb{S}=$ unit quaternion.)

Show that $G_{E}=\mathbb{S}^1\times \mathbb{S}^1$

Firstly, giving terminology, $M$ (or is oriented $2-$planes in $\mathbb{R}^4$) is a set of equivalence classes of pair of two orthonormal elements $(v,w)\in \mathbb{S}^3\times \mathbb{S}^3$ where the $(v_1,v_2)\sim (w_1,w_2)$ if they are related by an element in $SO(2)$. (we can also think as $(v_1,v_2 )\sim(w_1,w_2)$ $\iff $ the two pairs are in same oriented plane.)
And $G_E$ is an isotropy subgroup(or stabilizer) of group action of $\mathbb{S}^3\times \mathbb{S}^3$ on $M$ at $E=[(e_1,e_2)]$ where $$e_1=\begin{bmatrix} 1\\0\\0\\0  \end{bmatrix}\hspace{4mm}e_2=\begin{bmatrix} 0\\1\\0\\0\end{bmatrix}.$$
Here, the action of $\mathbb{S}^3\times \mathbb{S}^3$ on $M$ is induced action by a map $f:\mathbb{S}^3\times \mathbb{S}^3 \rightarrow SO(4)$ such that$$(p,q)\cdot [(v,w)]=f(p,q)\cdot [(v,w)] $$
where $f(p,q)\cdot v=pvq^{-1}$ for any $v\in \mathbb{R}^4=\mathbb{H}$. The multiplication is quaternion multiplication. It might be helpful to refer this 'The Quaternions and $SO(4)$'
In this regards, we can consider $\mathbb{S}$ as a unit quaternion group. 
I am thanking for any hint or answer in advance.  
 A: By definition, $G_E = \{(p,q)\in S^3\times S^3: [(p\,e_1\, q^{-1}, p\,e_2\, q^{-1})] = [(e_1, e_2)]$.
Here's a computational way to figure out begin to figure $G_E$:  Write $p = p_0 + p_1 i + p_2 j + p_3 k$ and similarly for $q$.
Then $p \, e_1 q^{-1} = pq^{-1} \in \operatorname{span}_{\mathbb{R}}\{1,i\}$.  In particular, the $j$ and $k$ components of $pq^{-1}$ must vanish.
Noting that $q^{-1} = \frac{\overline{q}}{|q|^2} = \overline{q}$, we see that the vanishing of the $j$ and $k$ components of $pq^{-1}$ is equivalent to \begin{eqnarray} -p_0 q_2 + p_1 q_3 + p_2 q_0 - p_3 q_1 &= 0\\
-p_0q_3 - p_1 q_2 + p_2 q_1 + p_3 q_0 &=0\end{eqnarray}
Likewise, the fact that $p\, e_2 \, q^{-1} = piq^{-1}$ has no $j$ and $k$ components is equivalent to \begin{eqnarray} p_0q_3 + p_1 q_2 + p_2 q_1 + p_3 q_0 &=0 \\-p_0 q_2 + p_1 q_3 - p_2 q_0 + p_3 q_1 &= 0 \end{eqnarray}
We can view these four equations together as a linear system where the coefficents are the $qs$ and the variables are $p_0, p_1, p_2, p_3$.  In other words, this is equavalent to $$\begin{bmatrix} -q_2 & q_3 & q_0 & -q_1\\ -q_3 & -q_2 & q_1 & q_0 \\ q_3 & q_2 & q_1 & q_0\\ -q_2 & q_3 & -q_0 & q_1  \end{bmatrix} \begin{bmatrix} p_0\\ p_1 \\ p_2 \\ p_3\end{bmatrix} = 0.$$
Now the $P$ vector is non-zero because $|p| = 1$, as $p\in S^3$.  Thus, the $Q$ matrix must have zero determinant in order to have a solution to this matrix equation.  The determinant (according to Wolfram alpha) is $ 0 =4(q_0^2 + q_1^2)(q_2^2 + q_3^2)$
Thus, either $q_0 = q_1 = 0$ or $q_2 = q_3 = 0$.  We next actually show the first case cannot occur.  (If you aren't worried about the oriented $(1,i)$ plane, then both cases do arise).
So, assume $q_0 = q_1 = 0$.  Then the above matrix equation forces $p_0 = p_1 = 0$ as well.  That is, $q = q_3 j + q_4 k$ and similarly for $p$.  Note that $p_3^2 + p_4^2 = q_3^2 + q_4^2 = 1$ because they are just $|p|^2 = |q|^2 = 1$ in disguise.
Now, compute $$p\, e_1\, q^{-1} = pq^{-1} = (p_3 q_3 + p_4 q_4) + (-p_3 q_4 + p_4 q_3)i$$ and likewise $$p\, e_2\, q^{-1} = piq^{-1} = (-p_3 q_4 + p_4 q_3) + (-p_3 q_3 - p_4 q_4)i.$$
It follows that the matrix of the transformation $e_n \mapsto p e_n \, q^{-1}$ is given by $\begin{bmatrix} p_3 q_3 + p_4 q_4 & -p_3 q_4 + p_4 q_3\\ -p_3 q_4 + p_4 q_3 & -p_3 q_3 - p_4 q_4 \end{bmatrix},$ which has determinant $-(p_3^2 + p_4^2)(q_3^2 + q_4^2) =-1$ (with thanks, again, to Wolfram alpha).
In other words, when $q_0 = q_1 = 0$, we get an orientation reversing map of the plane, so it doesn't map the oriented plane to itself.  This gets rid of that case.
Thus, we are left with the case that $q_2 = q_3 = 0$.  In other words, $q = q_0 + q_1 i \in S^1\subseteq S^3$.  It's not too hard to see using the $4\times 4$ matrix above that $p_2 = p_3 = 0$, so $P$ has this form.  All this shows that $G_E\subseteq S^1\times S^1$.
Thus, to conclude the problem, we must show that $S^1\times S^1\subseteq G_E$.  Or, said another way, for any element of the form $(e^{i\theta}, e^{i\phi})\in S^3\times S^3$, that $e^{i\theta} P q^{-i\phi} = P$ where $P$ is the oriented plane $\operatorname{span}\{e_0, e_1\}$.  I will leave that calculation to you.
