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Let $Q$ be the quaternion group where $|\operatorname{Aut}(Q)|=24$ and consists of elements $\phi$ with $\phi(i)\in \left \{ \pm i, \pm j,\pm k \right \}$ and $\phi(j)\in \left \{ \pm i, \pm j,\pm k \right \}\setminus \left \{ \pm\phi(i)\right \}$.

Prove that $\operatorname{Aut}(Q)$ contains no element of order $6$, and so $\operatorname{Aut}(Q)\simeq S_{4}$.

Any help is appreciated. I have seen the other posts, but I am not sure how to approach this proof showing $\operatorname{Aut}(Q)$ contains no element of order $6$, then to conclude that it is isomorphic to $S_4$. There is a link in another post, that refers to no element of order $6$, but unable to be accessed.

user565684
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  • The post indicated is very helpful, but not sure how to relate the details on proving $Aut(Q)$ contains no element of order $6$. – user565684 Nov 06 '18 at 09:00
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    An element $\alpha$ of order $6$ would induce a $3$-cycle on the three subgroups of order $4$, and since $\alpha^3 \ne 1$, it would have to map all elements $x$ of order $4$ to $x^{-1}$. But this is impossible since, if $Q =\langle x,y \rangle$ then $x^{-1}y^{-1} \ne (xy)^{-1}$. But I am not sure how this helps prove ${\rm Aut}(Q) \cong S_4$. Perhaps you need to prove that $|{\rm Aut}(Q)|=24$ first? – Derek Holt Nov 06 '18 at 11:14
  • Derek Holt's answer to the order 6 question is great! However I second his comment that it is completely unclear how 'no element of order 6' implies $S_4$. Where did you find this problem? Did the source comes with a list of all groups of order 24 so that after showing no element of order 6 you can eliminate most of them on that basis? To me the general problem of finding all 24 element groups and seeing which ones have elements of order 6 sounds MUCH harder than the very specific problem of finding $Aut(Q_8)$. – Vincent Nov 06 '18 at 11:30

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