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.

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    $\begingroup$ 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 $SL(2,3)$? $\endgroup$ Commented Apr 10, 2021 at 9:03
  • $\begingroup$ @DietrichBurde No we don't have this information in the book. But yes it would be intersesting to use the equivalent presentation. $\endgroup$
    – acd3456
    Commented Apr 10, 2021 at 9:09
  • $\begingroup$ 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 $2$. $\endgroup$ Commented Apr 10, 2021 at 14:44
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    $\begingroup$ 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. $\endgroup$ Commented Apr 10, 2021 at 15:08
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    $\begingroup$ @ancientmathematician Ok I see the trick, thanks. $\endgroup$
    – acd3456
    Commented Apr 10, 2021 at 15:16

1 Answer 1


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

  • $\begingroup$ Can you explain, in the commutator line, how do you get the fourth equality ? I can't get over it. $\endgroup$
    – acd3456
    Commented Apr 11, 2021 at 11:35
  • $\begingroup$ $t^2= t^{-1}u$ and $s^2=s^{-1}u$, so $t^2s^2t^2 = u^{3}t^{-1}s^{-1}u^{-1}$. $\endgroup$
    – Derek Holt
    Commented Apr 11, 2021 at 11:50

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