# 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?

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?

• I don't know whether we (you or I) can deduce that it is isomorphic to the quaternion group. I have not tried. But I generally trust computer algebra systems like GAP with problems like this, and they can indeed deduce that it is quaternion of order 8. I would guess that it is not too hard to do it by hand. – Derek Holt Feb 24 at 18:16

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.
• Very clean! ${}$ – Jyrki Lahtonen Feb 24 at 19:33
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.
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.