I was reading a book on quaternion algebras and found the following exercise and was unable to do it and it has been really frustrating me, Let $(a,b)_F$ be a quaternion algebra with $i^2=a$ , $j^2=b$ and $ij=-ji$
Prove that $(a,b)_F$ , $(b,a)_F$ and $(ac^2,b)_F$ are isomorphic and prove that $(a^2,b)_F$ is not a division algebra.
Let's show that $(a,b)_F$ and $(b,a)_F$ are isomorphic. Let's say $(a,b)_F$ is generated by $i,j$ with
$$ i^2=a, \quad j^2=b, \quad ij=-ji $$
and $(b,a)_F$ is generated by $k$ and $\ell$ satisfying
$$ k^2=b, \quad \ell^2=a, \quad k\ell=-\ell k. $$
To define an isomorphism $\phi:(b,a)_F\to(a,b)_F$ it suffices to specify where $\phi$ sends the two generators $k$ and $\ell$. We must send $k$ and $\ell$ to two elements of $(a,b)_F$ where $\phi(k)$ squares to $b$, $\phi(\ell)$ squares to $a$, and $\phi(k)$ and $\phi(\ell)$ anticommute. Can you see two such elements of $(a,b)_F$ to send $k$ and $\ell$ to with $\phi$?
For showing $(a^2,b)_F$ is not a division algebra, try and find a nonzero element that is not invertible; one way this may occur is with zero divisors. Look at the new relation $i^2=a^2$. What can you do to this equation to exhibit zero divisors?