2
$\begingroup$

Is there a way to find the group whose presentation is given by $\langle a, b, c \mid a^{2} = b^{2} = c^{2} = abc = e, ab = ba, ac = ca, bc= cb\rangle$?

  • 1
    Perhaps interesting to note that you don't even need the "commuting" relations; $a^2=b^2=c^2=abc=e$ is already enough to deduce the structure of the group.2011-11-03

1 Answers 1

7

The last three relations say that the generators commute, so the group is abelian, and every element has the form $a^ib^jc^k$. Furthermore, since $a^2=b^2=c^2=e$ you can reduce each of the exponents modulo $2$. Thus without the $abc=e$ relation the group is the additive group of the vector space $(\mathbb F_2)^3$. The last relation then kills the subspace generated by $abc$, and we end up with something isomorphic $(\mathbb F_2)^2$, which is just $V_4$.