Automorphisms of Quaternion Group $Q$. Prove that $\operatorname{Aut}(Q)$ contains no element of order $6$, and so $\operatorname{Aut}(Q)\cong S_4$ [duplicate]

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.

marked as duplicate by Alexander Gruber♦ abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 7 '18 at 7:17

• 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 6 '18 at 9:00
• 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 6 '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 6 '18 at 11:30