In the book An introduction to the theory of groups, by Rotman the author gives us the presentation of the binary tetrahedral group
$$B = \langle r,s,t \mid r^2 = s^3 = t^3 = rst \rangle.$$
After that in the exercise 11.13 he asks us to show that the generator $s$ has order $6$ in $B$.
My question is how can I do that without knowing that $B$ is a subgroup of quaternions' units and by only using the presentation ? Because otherwise it's just an easy calculation.
It can be checked that there is a homomorphism $\phi:B\to {\rm SL}(2,3)$ with $$\phi(r) = \left(\begin{array}{cc}0&1\\2&0\end{array}\right),\ \ \phi(s) = \left(\begin{array}{cc}2&0\\2&2\end{array}\right),\ \ \phi(t) = \left(\begin{array}{cc}0&2\\1&1\end{array}\right),$$ with $\phi(r^2)=\phi(s^3)=\phi(t^3) = \phi(rst) = -I_2$, in which $\phi(s)$ has order $6$, so we just need to check that $s^6=1$ in $B$.
The relations imply that $r=st$, and the element $u=rst$ is central, with $s^3=t^3=(st)^2=u$. Let $x=st$ and $y=ts$.
Then $x^2=u$ and, since $u$ is central, $y^2=tx^2t^{-1}=u$.
Then (using $t^{-1}s^{-1}t^{-1}=su^{-1}$), we have $$[x,y] = x^{-1}y^{-1}xy=u^{-2}xyxy = u^{-2}st^2s^2t^2s = ust^{-1}s^{-1}t^{-1}s = s^3 = u,$$ so, since $[x,y]$ is central, we have $u^2=[x,y]^2=[x^2,y]=[u,y]=1,$ and so $s^6=1$ as claimed.
(Note that $\langle x,y \rangle \cong Q_8$.)