It's $\mathbb{H}$ isomorphic to $\mathbb{H^{op}}$? Let $\mathbb{H}$ be the quaternions division ring and  $\mathbb{H^{op}}$ the opposite ring.  The additive structure  is the same but the multiplication is given by: $x\star y = y\cdot x,\forall x, y\in \mathbb{H}$, where '$\cdot$' is the usual multiplication. It's $\mathbb{H}$ isomorphic to $\mathbb{H^{op}}$ as division rings?
 A: Isomorphisms $R\cong R^{\mathrm{op}}$ of a ring $R$ correspond to antiautomorphisms $\alpha$ (note this means that $\alpha$ satisfies the property $\alpha(rs)=\alpha(s)\alpha(r)$ instead of the usual rule). Let $(R,+,\cdot)$ be a ring with underlying set $R$ and define $\star$ by $a\star b=b\cdot a$. Then $(R,+,\star)$ is the opposite ring. The condition that an additive map $\alpha:R\to R$ satisfy $\alpha(r\cdot s)=\alpha(s)\cdot \alpha(r)$ (i.e. that $\alpha$ is an antiautomorphism) is equivalent to $\alpha(r\cdot s)=\alpha(r)\star\alpha(s)$ (i.e. that $\alpha$ defines an isomorphism between the ring and its opposite ring, $(R,+,\cdot)\to(R,+,\star)$.)
The quaternions $\mathbb{H}$ have conjugation as an antiautomorphism.
A: A short argument involving the Brauer group goes as follows: the reason why the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order $2$ in the Brauer group of $\mathbb{R}$, and hence equal to its own inverse in the Brauer group. 
