Is $\operatorname{Aut}(D_{12})\simeq D_{12}$?

Question

Let $D_{12}$ be the dihedral group of order 12. Then $$|\operatorname{Aut}(D_{12})|=6\phi(6)=12=|D_{12}|,$$ and the standard method of proof for $$\operatorname{Aut}(D_6)\simeq D_{6}\qquad\mbox{and}\qquad \operatorname{Aut}(D_8)\simeq D_{8}$$ seems to also work for $$\operatorname{Aut}(D_{12})\simeq D_{12}.$$ But in this article (p.461, 14th line from the top), it says that $n=3$ and $n=4$ are the only numbers for which $$\operatorname{Aut}(D_{2n})\simeq D_{2n}.$$ Could this be an error? Or am I missing something?

You are right, $\mathrm{Aut}(D_{12})\cong D_{12}$. I will look at the paper. — David A. Craven, Jul 29 '20 at 08:53
You are right, and it has been shown here at this post, where the notation $D_6$ is used for the dihedral group having $12$ elements. — Dietrich Burde, Jul 29 '20 at 09:49

score 3 · Accepted Answer · answered Jul 29 '20 at 08:59

The article is wrong. First, the automorphism group is definitely $D_{12}$: here is Magma code that proves it.

> G:=DihedralGroup(6);
> A:=AutomorphismGroup(G);
> A:=PermutationGroup(A);
> IdentifyGroup(A);
<12, 4>
> IdentifyGroup(G);
<12, 4>

Second, even his Theorem A states that $\mathrm{Aut}(D_{12})\cong \mathrm{Hol}(Z_6)$, and this is dihedral of order $12$.

Is $\operatorname{Aut}(D_{12})\simeq D_{12}$?

1 Answers1

Linked