# Automorphism of $Q_8$ [duplicate]

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!

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

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.