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This question already has an answer here:

Is there anyone could help me to prove that $Aut(Q_8)=S_4$?

Someone told me that there's an isomorphism between the rigid motions of cube and $Aut(Q_8)$, any ideas?

Thank you!

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marked as duplicate by Derek Holt, Davide Giraudo, Trevor Wilson, TZakrevskiy, Lord_Farin Oct 8 '13 at 19:17

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    $\begingroup$ This could be a good place to have a look at. $\endgroup$ – onimoni Oct 8 '13 at 17:47
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In $Q_8$ the central elements $1,-1$ are clearly fixed by all automorphisms. The remaining thr pairs $(i,-i)$, $(-j,-j)$, $(k,-k)$ must also stay together (they could be permuted among each other, and within a pair the elements could be swtched). It is natural to associate these pairs with the three pairs of opposite faces of a cube. Associating $i,j,k$ to faces such that the vectors from the center to them form a positively oriented reference frame, try to show that all positive reference frames correspond to oriented triples $(x,y,z)$ with $xy=z$. It is clear that automorphisms of $Q_8$ map such triples into each other; if you show that they do so (simply) transitively, then you are done.

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