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Let $G$ be a group with elements $\{e, a, b, c, \theta, \theta a, \theta b, \theta c \}$ where $a^2 = b^2 = c^2 = \theta$, $\theta^2 = e$, $ab = \theta b a = c$, $bc = \theta c b = a$, $ca = \theta a c = b$.

Can you find an easier-to-handle group isomorphic to $G$? (say a subgroup of $S_n$)

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    $\begingroup$ The smallest such $n$ is $n=8$, so that's not very helpful! It can be written as group of $2 \times 2$ matrices over the complex numbers. $\endgroup$
    – Derek Holt
    Commented May 4, 2013 at 21:58
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    $\begingroup$ That looks like the quaternion group. $\endgroup$ Commented May 4, 2013 at 22:23

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The group you describe is the quaternion group $Q_8$. Writing $\theta$ as $-1$ makes these relations a little easier to work with, as this notation emphasizes implicitly that $\theta$ commutes the rest of the group. $$a^2=b^2=c^2=-1\\ \begin{array}{ccccccc}ab=-ba=c&&&bc=-cb=a&&&ca=-ac=b\end{array}$$ Usually you see $a,b,$ and $c$ written $i,j,$ and $k$, respectively. It can be shown that it is equivalent to write these relations as $$(-1)^2=1\\i^2=j^2=k^2=ijk=-1.$$ To see $Q_8$ as a permutation group, we use Cayley's theorem to map its generators into $S_8$. Since $i^2=j^2=-1$ and $k=ij$, it suffices to find the images of $i$ and $j$, from which we may compute $$ \begin{array}{rccccrcc} 1&\mapsto&\text{id}&&-1&\mapsto&(16)(24)(38)(57)\\ i&\mapsto&(1264)(3587)&&-i&\mapsto&(1462)(3785)\\ j&\mapsto&(1765)(2843)&&-j&\mapsto&(1567)(2348)\\ k&\mapsto&(1863)(2547)&&-k&\mapsto&(1368)(2745) \end{array} $$

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  • $\begingroup$ Nice answer +1. $\endgroup$
    – Mikasa
    Commented May 5, 2013 at 8:19
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There are only two nonabelian groups of order 8, namely $D_8$ and $Q$, the quaternion group. Your group is evidently $Q$, with $(\theta, a,b, c) = (-1, i, j, k)$.

Cayley's theorem gives a canonical isomorphism between your group and a subgroup of $S_8$.

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