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I am trying to prove that $Aut(D_8) \equiv D_8$. It is not hard to see that $\lvert Aut(D_8)\rvert = 8$. Indeed, it is at most $8$ as $r$ (canonical rotation) has order $4$ and $s$ (canonical reflection) has order $2$. On the other hand, $D_8$ is normal in $D_{16}$ which acts by conjugation on $D_8$. So, the order is exactly $8$. However, I am having troubles to show that $Aut(D_8) \equiv D_8$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).


marked as duplicate by Dietrich Burde abstract-algebra Dec 12 '18 at 9:30

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    $\begingroup$ You can do it by showing that there are $f,g\in \text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$. $\endgroup$ – user9077 Dec 12 '18 at 0:55

You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.


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