If $$G=\langle a,b:a^8=b^2a^4=ab^{-1}ab=e\rangle,$$
how can I prove that $G$ has order at most $16$?
I have played with the relations for a while, but am literally stuck. I know that the order of $b$ has to be at most equal to $4$, and I also found that $aba=b$, but I don't really know how to connect these relations in order to prove that $G$ has at most $16$ elements. I am thinking that maybe Von Dyck's theorem might be useful. I found this article http://buzzard.ups.edu/courses/2012spring/projects/clausen-groups-16-ups-434-2012.pdf and don't know which might be the group I should be looking at in order to use Von Dyck's theorem, or if there is another way to approach this problem.
Any help is much appreciated.
This group is the generalized quaternion group $Q_{16}$ of order $16$. Try to show that the only (pairwise distinct) elements of $G$ are:
$1, a, b, a^2, b^2, ab, a^3, ab^2, a^2b^{-1}, b^{-1}, a^{-2}, a^{-1}b, ab^{-1}, a^{-1}, a^2b, a^{-1}b^{-1}$.