# Order of a generator with group presentation

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.

• See wikipedia, we have $s^3=-1$, so $s$ has order $6$. In Rotman's book, can we already use that $B$ is isomorphic to $SL(2,3)$? Commented Apr 10, 2021 at 9:03
• @DietrichBurde No we don't have this information in the book. But yes it would be intersesting to use the equivalent presentation. Commented Apr 10, 2021 at 9:09
• Does it help to show that there is a surjection of $B$ onto $A_4$ whose kernel is the (central) element $rst$? I think that then forces $\langle rst\rangle$ to have order $1$ or $2$. Commented Apr 10, 2021 at 14:44
• Choose any element of order $3$ in $A_4$ as image of $s$, wlog $(123)$. Now what shall we choose for image of $t$? - well there's bound to be an overlap of two elements so might as well try $(124)$ wlog. Then the product is $(14)(23)$ which has order $2$ and so it all works. Commented Apr 10, 2021 at 15:08
• @ancientmathematician Ok I see the trick, thanks. Commented Apr 10, 2021 at 15:16

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 $$\langle x,y \rangle \cong Q_8$$.)

• Can you explain, in the commutator line, how do you get the fourth equality ? I can't get over it. Commented Apr 11, 2021 at 11:35
• $t^2= t^{-1}u$ and $s^2=s^{-1}u$, so $t^2s^2t^2 = u^{3}t^{-1}s^{-1}u^{-1}$. Commented Apr 11, 2021 at 11:50