I think that the presentation of $Q_8$ is $\langle i,j:i^2=j^2,ij(-i)=-j\rangle$, but in many text books it given an extra condition $i^4=1$. My question is that how we can derive this extra condition from prevous conditions.
My main problem is: I am stydying a research article (https://www.sciencedirect.com/science/article/pii/0021869387902481) . In which this $Q_{4n}=\langle x,y:x^2=y^n,x^{-1}yx=y^{-1}\rangle$ presentation of $Q_{4n}$ is given. I do't know how we can conclude that order of $Q_{4n}$ is $8n$ and how order of $y$ is $2n$.
I tried a lot but only I can show that $y^{4m}=1$, please help me.
I'll show that $i^4=1$ in the quaternion group, you should be able to take it from there.
Note that $i^4=i^2j^2=i(ij)j$ since $i^2=j^2$.
The relations $ij(-i)=-j$ can be written as $ij=j^{-1}i$. I will not write $-i$ for the inverse of $i$ as this notation suggests structure other than just group structure!
Using the relation we find that $i^4=i(ij)j=i(j^{-1}i)j=ij^{-1}(ij)=ij^{-2}i$. Since $i^2=j^2$, the inverses are equal as well, i.e. $j^{-2}=i^{-2}$. Hence $i^4=ij^{-2}i=ii^{-2}i=e$ yields the neutral element.