About the set of pure Quaternion I have a question about the set of pure Quaternion:
Let $\mathbb K$ be a field of $\text{Char}$($\mathbb K$)$\neq 2$.
Set $Q=Q(a,b\mid \mathbb K)=(a,b)_{\mathbb K}$ be the Quaternion algebra for $a,b \in \mathbb K$, with $\mathbb K$ basis $1, i, j$ and $k$, such that
\begin{cases} i^2=a,\\ j^2=b,\\ k^2=-ab,\\ ij=-ji=k, \\ki=-aj, \\ ik=aj,\\ kj=bi,\\ jk=-bi \end{cases}
Then the set $Q_0=Q_0(a,b\mid \mathbb K)=\langle \mathbb K i+\mathbb Kj+\mathbb Kk \rangle $ with the scalar part equal to zero will be called the set of pure Quaternion.
Now the question is as follows:


(1) Show that $x \in Q_0$ if and only if $x\notin \mathbb K$ and $x^2 \in \mathbb K$ and conclude that if $\varphi: Q \to Q'$ is a $\mathbb K$-linear isomorphism of such quaternion algebras, then $\varphi (Q_0)=Q_{0}^{'}$.
    (2) Let $[,]:Q \times Q \to Q$ be the commutator, i.e. $[x:y] = xy - yx$ for all $x,y \in Q.$ Show that $[Q_0,Q_0]\subset Q_0$ with equality if $a\neq 0 \neq b.$
    (3) Let $a,b \in \mathbb K- \{0\}$. Show that there $Q(a,b\mid \mathbb K) \simeq Q(b,a\mid \mathbb K)\simeq Q(a,-ab\mid \mathbb K)$ (as $\mathbb K$-algebras)


For the first part, we have that for $q\in Q$, $q=x_0+q_0$ for $x_0 \in \mathbb K$ and $q_0 \in Q_0$. Then we claim that $q^2\in \mathbb K \Leftrightarrow q\in \mathbb K ~\text{or} ~q ~ \text{is pure}$.
We see that $q^2=(x_0+q_0)^2=x_{0}^{2}+2x_0q_0+q_{0}^{2}$ for $x_0\in \mathbb K$ and $q_0\in Q_0$. And as we know $q_{0}^{2}=(x_1i+x_2j+x_3k)^2=-x_{1}^{2}-x_{2}^{2}-x_{3}^{2} \in \mathbb K$. So for to prove that $q^2\in \mathbb K$, we need to show that $x_0q_0 \in \mathbb K$. Which means that $q_0 \in \mathbb K$ or $x_0=0$. And this means that $q\in \mathbb K$ or $q$ is pure. And we can conclude that the pure quaternions in $Q$ are precisely the $q$ satisfying $q^2 \in \mathbb K$ with $q \notin \mathbb K$ along with 0.
And $\varphi: Q \to Q'$ is a $\mathbb K$-linear isomorphism of such quaternion algebras if it takes center to center and we know that $\mathbb K$ is in its center and if for $q\in Z(Q)$, $q\in \mathbb K$ then $q^2\in \mathbb K$ and then by above we can conclude that $q$ is pure or an scalar. So we have $\varphi (Q_0)=Q_{0}^{'}$.
For the second part, let $q_1=ix_{11}+jx_{12}+kx_{13}$ and $q_2=ix_{21}+jx_{22}+kx_{23}$ be in $Q_0$. Then $q_1 q_2= bi(x_{13}x_{22} - x_{12}x_{23} +aj(x_{11}x_{23} -x_{13}x_{21})+k(x_{11}x_{22}-x_{12}x_{21})+ax_{11}x_{21}+bx_{12}x_{22}-abx_{13}x_{23}$ and $q_2q_1=-aj(x_{23}x_{11}-x_{21}x_{13})-bi(x_{22}x_{13}-x_{23}x_{12})-k(x_{22}x_{11}-x_{21}x_{12})+ax_{21}x_{11}+bx_{22}x_{12}-abx_{23}x_{13}$ and so we have $q_1 q_2-q_2q_1=2bi(x_{13}x_{22} - x_{12}x_{23} +2aj(x_{11}x_{23} -x_{13}x_{21})+2k(x_{11}x_{22}-x_{12}x_{21})$ which is an element of $Q_0$ if $a\neq 0\neq b$.
Please let me know if my calculations and logic in proving the first two statements is not correct?
Thanks!
For the third part, I do not know how to show it.
Can someone let me know how to show it?
Thanks!
 A: 
So for to prove that $q^2\in \mathbb K$, we need to show
  that $x_0q_0 \in \mathbb K$.

I would say: "So $q^2\in \mathbb K$ is equivalent to $x_0q_0 \in \mathbb K$."

Which means that $q_0 \in \mathbb K$ or
  $x_0=0$.

Or: Since $x_0q_0 \in Q_0$ and $\mathbb{K}\cap Q_0 = \{0\}$, by definition, $x_0q_0 = 0$ which is equivalent to "$x_0=0$ or $q_0=0$". The rest seems fine until

And $\varphi: Q \to Q'$ is a $\mathbb K$-linear isomorphism of such
  quaternion algebras if it takes center to center and we know that
  $\mathbb K$ is in its center and if for $q\in Z(Q)$, $q\in \mathbb K$
  then $q^2\in \mathbb K$ and then by above we can conclude that $q$ is
  pure or an scalar. So we have $\varphi (Q_0)=Q_{0}^{'}$.

I don't understand that. I would think: since $\varphi$ is a $\mathbb{K}$-algebra iso, for $q\in Q$, we have $q^2\in \mathbb{K} \Leftrightarrow \varphi(q)^2 =\varphi(q^2) \in \varphi (\mathbb{K}) = \mathbb{K}$ as well as $q\in \mathbb{K} \Leftrightarrow \varphi(q)\in \mathbb{K}$, and so ...
For the second part, your computation looks correct, and certainly the result is in $Q_0$, by the way regardless of whether $a,b$ are $=0$ or not. It might be smart to shorten the computation: Using that $[ , ]$ is bilinear and $[x,y] = -[y,x]$, it would suffice to show that $[i,j], [i,k]$ and $[j,k]$ are in $Q_0$.
But you still have to show that $[Q_0,Q_0] = Q_0$ if $a\neq 0\neq b$. For that, by linearity it would suffice to find certain elements $x_i,y_i \in Q_0$ such that $[x_i,y_i]=i$, and similarly for $j$ and $k$. The computations above should give you a hint, and probably you have to invert $a$ and $b$ somewhere (because we have to use $a\neq 0 \neq b$ somewhere).
The third part has little to do with $Q_0$. You'll have to write down explicit isomorphisms. The first looks like maybe "switching" $i$ and $j$ (i.e. the $i$ of the first algebra gets sent to the $j$ of the second, and vice versa) should do it, but be careful with the signs. For the second iso, maybe some other pair gets switched (and maybe something multiplied with $-1$)?
