I was playing around in GAP and I noticed an interesting pattern when computing $|\mbox{Aut} (Q_{2^n})|$ for different $n$. For $n=3$, I am aware that $\mbox{Aut}(Q_8) \cong S_4$, however for $n > 3$, the pattern seems to be $$|\mbox{Aut} (Q_{2^n})| = 2^{2n - 3}.$$ However, I only checked $n$ up to 9. My question is whether the formula is valid for $n > 3$, and if so, is there an easy way to see this?
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1Relevant: $$\operatorname{Aut}(Q_{2^n})\cong \operatorname{Aut}(D_{2^n}).$$ – Shaun Sep 13 '24 at 12:32
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1Combining the above with this should do the trick. – Shaun Sep 13 '24 at 12:34
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To summarise my comments into an answer:
Showing ${\rm Aut}(D_{2^n})\cong{\rm Aut}(Q_{2^n})$ for $n\ge 4$.
and from here:
What is the automorphism group of $D_5$ (and, more generally, of $D_n$)?
$$\begin{align} |\operatorname{Aut}(D_{2^{n-1}})|&=2^{n-1}\varphi (2^{n-1})\\ &=2^{n-1}\times 2^{n-2}\\ &=2^{2n-3} \end{align}$$
for $n>3$.
Note: this is with the convention $|D_m|=2m$, so you're correct!
Shaun
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