Showing $a^2=b^2$ in a group with 8 elements isomorphic to Quaternion group I am working in a group $G$ with 8 elements such that $a, b\in G$ and $ba=a^3b$. Moreover, I am given that both $b$ and $a$ have order 4 and $\langle a \rangle \neq \langle b \rangle$. I understand this group is isomorphic to Quaternion group, but this is not given and I am trying to prove this. I want to show $a^2=b^2$, but I am stuck.
Here is what I have done so far:
$$(ab)^2=abab=a(ba)b=aa^3bb=a^4b^2=b^2$$
Also from the equality $ba=a^3b$ I can derive many other equalities, but they do not seem to give me anything useful. For instance
$$aba=b \text{ or } a^3ba^3=b$$
Another observation I have made is that both $b^2$ and $a^2$ have order 2. But I am not sure if that is enough to conclude $a^2=b^2$. Please guide me in the right direction.

Similar to @GerryMyerson's comment, use the fact that there are only two nonabelian groups of order $8$.
For instance, the quaternions only have one element of order $2$.  The dihedral group has $5$.
