Show that $SU(2)\times SU(2)/\ker(\varphi)\cong SO(4)$ I want to show that $SU(2)\times SU(2)/\ker(\varphi)\cong SO(4)$ where $\varphi:SU(2)\times SU(2)\rightarrow \rm End(\mathbb{H})$ is the map $\varphi(A,B)=(h\mapsto AhB^{-1})$. 
I have shown that $\ker(\phi)={(I,I),(-I,-I)}$, and $\varphi(SU(2)\times SU(2))\subset SO(4)$. 
I have tried to use the isomorphism theorem, but I am getting stuck. To use the isomorphism theorem, I want to show that for any $X\in SO(4)$ there is $A, B$ such that $\varphi(A, B)=X$. This gives a rather horrible matrix expression that I cannot solve. Is there a slicker way to approach this?
 A: Yes. The Cartan-Dieudonne theorem says every rotation in $\mathbb{R}^n$ is a composition of (an even number less than or equal to $2n$) reflections. Notice $F(x)=-\overline{x}$ flips the real component while keeping the imaginary components fixed, so it is a reflection across the $3$D space of purely imaginary quaternions, or equivalently a reflection across the orthogonal complement $1^{\perp}$. To get a reflection $F_q$ across any other orthogonal complement $q^{\perp}$, we may "transport" the one from $1^{\perp}$: first rotate $q$ to $1$, then reflect across $1^{\perp}$, then rotate $1$ back to $q$. The rotation from $1$ to $q$ may be left or right multiplication by $q$ (and rotating back the same but by its inverse $q^{-1}=\overline{q}$, assuming $q$ is a unit quaternion). That is,
$$ F_q(x) ~=~ q\big(-\overline{\overline{q}x}\,\big) ~=~ \big(-\overline{x\overline{q}}\,\big)q ~=~ -q\overline{x}q. $$
Composing two of them gives $(F_p\circ F_q)(x)=(pq)x(qp)$, which is of the form $axb$. Therefore, the group homomorphism $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to \mathrm{SO}(4)$ must be onto.
(People often use $\mathrm{SU}(2)$ to refer to $\mathrm{Sp}(1)$ since they're isomorphic, for basically the same reason that $SO(2)$ is isomorphic to $U(1)$. But people don't insist on referring to complex numbers as real matrices even though it can be done. The reason for the quaternion-phobia, I expect, is the idea that complex matrices are supposed to be more familiar than quaternions. Here, $\mathrm{Sp}(1)$ refers to the $1\times 1$ quaternionic unitary matrices, which is essentially the group $S^3$ of unit quaternions, just as $U(1)$ is essentially $S^1$.)
To see the kernel is $(-1,-1)$, first suppose $axb=x$ for all $x$, set $x=1$ to see $b=a^{-1}$, then assuming $a$ has nonzero imaginary part let $x$ be a perpendicular imaginary quaternion in which csae $axa^{-1}=x$ simplifies to the expression $a^2x=x$, and $a^2=1$ implies $a=\pm1$.
