If we are handed the presentation $\langle i,j,k \mid i^2=j^2=k^2=ijk \rangle$ and nothing more, can we deduce that this is the quaternion group?

Question

If we are handed the group presentation $\langle i,j,k \mid i^2=j^2=k^2=ijk \rangle$ and nothing more, can we deduce that this is the quaternion group?

Nothing in this presentation tells us that $i^2=j^2=k^2=ijk=-1$ and that $i^4=j^4=k^4=(ijk)^2=1$. Can we conclude these relations from the relation given in the presentation?

Answer 1

Yes, we can do it.

The cancellation laws immediately get us $i=jk$ and $k=ij$. Multiply the first by $i$ on the right, and $i^2=jki$, leading to $j=ki$ and completing that cycle.

Now, applying these laws $j^2=i^2$, $ij=k$, $jk=i$, we have the following chain of equalities: $$j^4i=j^2i^3=j^2ij^2=j^2kj=jij=jk=i$$ Apply the cancellation law to that and $j^4=1$. From there, $i^4=k^4=(ijk)^2=1$ follow easily.

Answer 2

Repeatedly using the existence of inverses in groups gives $$ijk=k^2\implies ij=k,\,ijk=i^2\implies jk=i,\,kijk=k^3\implies kij=k^2=j^2\implies ki=j.$$Define $m:=k^{-1}ji=i^2$; we would normally call this $-1$. Since $m=i^2=j^2=k^2$, $m$ commutes with everything so $$ji=mk,\,ik=j^{-1}mk^2=mj,\,kj=k^2mi^{-1}=mi.$$Finally, $m^2=k^{-1}mkk^{-1}ji=k^{-1}ij$ is the identity.

Answer 3

For this particular group presentation there is a simple way to use cancellation to identify it. First define $\,u := ii = jj = kk = ijk\,$ which commutes with $\,i,j,k.\,$ Now $\,(ij)k = u = kk\,$ and using cancellation $\,ij=k.\,$ Similarly, $\,i(jk) = u = ii\,$ and $\,jk=i.\,$ Next, $\,(ki)j = k(ij) = kk = u = jj\,$ and using cancellation $\,ki=j.\,$ Next, $\, uk = (jj)k = j(jk) = ji.\,$ Similarly, we get $\,ui = kj, \, uj = ik.\,$ Finally, $\,uuk = uji = iiji = iki = ij = k,\,$ and thus $uu = 1.$ The group has eight elements $\,\{1,i,j,k,u,ui,uj,uk\}\,$ and isomorphic to the quaternion group.