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$\begingroup$

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

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    Since ab=ba, ac=ca, and bc=cb, the group is abelian. Since aa=bb=cc=e, it is generated by a bunch of (ok just three) elements of order 2. Check the abelian groups you know that are generated by elements of order 2. It should be the second one you try.2011-11-03
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    $ab=c^{-1}=c$ (since $c^2=e$ so $c=c^{-1}$. Looks like the Klein 4-group.2011-11-03
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    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

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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 $V_4$.