Automorphism of $D_6$ I need find an explicit way to express the group $Aut(D_6)$, and I have not idea how write this group, maybe this is an semidirec product of some groups but I don´t see this. thanks.
 A: Hint: To enumerate the automorphisms, recall that an automorphism is uniquely determined by how it acts on the generators of the group. Consider the presentation, $$D_{12}=\langle r,s \mid r^6=s^2=1, srs=r^{-1}\rangle = \{ 1,r,r^2,r^3,r^4,r^5,s,sr,sr^2,sr^3,sr^4,sr^5\}.$$ The subgroup $\langle r\rangle$ is the unique subgroup of index 2, which must therefore map onto itself under any automorphism; i.e., for any $\sigma \in Aut(D_{12})$, we must have $\sigma(r)=r^k$ for some $k$. Now, figure out which values of $k$ are actually possible here. By the same reasoning, as $s\notin\langle r\rangle$, we must have $\sigma(s) \notin\langle r\rangle$, so we must have $\sigma(s)\in\{s,sr,sr^2,sr^3,sr^4,sr^5\}$. Putting this together, you should be able to describe all the automorphisms of $D_{12}$ by identifying where they send $r$ and $s$. In particular this will give you the order of the group $Aut(D_{12})$.
From there, if you want to describe $Aut(D_{12})$ as an abstract group, what you can do is find a small generating set of automorphisms (from which the other automorphisms can be obtained by composition) and then try to identify the relations between these generating automorphisms.
A: Perhaps you've seen that the group of inner automorphisms of a group, $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$ isomorphic to $G/Z(G)$?
Now $Z(D_6) = \{e, r^3\}$, and it is not hard to show that $D_6/Z(D_6)$ is a non-abelian group of order $6$. So $\text{Aut}(D_6)$ has order at least (and a multiple of) $6$.
Any automorphism must send only non-identity central elements to non-identity central elements, which in this case means an automorphism of $D_6$ must send a reflection to a reflection (since the rotation of order $2$ is central).
Furthermore, $r$ can only go to $r$ or $r^5$, so there are at most $12$ possible automorphisms (any automorphism is completely determined by its action on the generators $r,s$ of $D_6$, and we have only $12$ possible choices).
If $r \mapsto r$ and $s \mapsto sr$ is an automorphism (can you prove this?), and you show it is an outer automorphism, then you've demonstrated $\text{Aut}(D_6)$ has order $12$.
You might ask yourself: what is the order of this automorphism? You might also ask, does it commute with the inner automorphism induced by $s$?
Can you tell now what $\text{Aut}(D_6)$ is?
