Computing $\operatorname{Aut}(D_{2n})$ and realizing it as $\mathbb{Z}/n\mathbb{Z} \rtimes_{\psi} (\mathbb{Z}/n\mathbb{Z})^{\times}$

Question

I have no idea how to approach this problem. I would like to find $\psi$ that does this. Given the fact that we know the presentation $D_{2n} = \langle x,y : x^n = e, y^2 = e,yxyx = e \rangle$.

Answer 1

We will take into account the following commutative diagram of homomorphisms. For readability I replaced $D_{2n}$ by $D$, the group of rotations by $C_n$ and the quotient group by this normal subgroup by $C_2$. $\iota$ is the natural injection of $C_n$ in $D$ , $\pi$ the natural projection of $D$ into the quotient $D/C_n = C_2$. $$ \begin{matrix} C_n & \stackrel{\iota}{\rightarrow} & D & \stackrel{\pi}{\rightarrow} & C_2 \\ \downarrow \alpha & &\downarrow \gamma & & \downarrow id \\ C_n & \stackrel{\iota}{\rightarrow} & D & \stackrel{\pi}{\rightarrow} & C_2 \\ \end{matrix} $$ Now if $\gamma$ is any automorphism of D then there is only one automorphism $\alpha$ that makes the diagram commutative. One can view $\alpha$ as the restriction of $\gamma$ to $C_n$. This gives us a homomorphism $res : Aut(D) \rightarrow Aut(C_n)$, the latter group being generated by an automorphism that maps $y \mapsto y$ and $x \mapsto x^d$ with $\gcd(d,n) = 1$. This group is isomorphic with the cyclic group of order $\Phi(n)$. The kernel of $res$ consists of the maps that map $y \mapsto yx^e \text{ with } 0\leq e \leq n-1$ and $x \mapsto x$. This group is isomorphic to the cyclic group of order $n$. So we have proven that $D = C_n \rtimes C_{\Phi(n)}$. To see the action of $\psi$ take a generator of $C_{\Phi(n)}$, i.e. a map $(x,y) \mapsto (x^d, y)$ and look at the action on the generating map $(x,y) \mapsto (x,yx)$ giving the map $(x,y) \mapsto (x, yx^d)$.

Answer 2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Suppose $n > 2$. Then $\Span{x}$ is the unique cyclic subgroup of order $n$, thus it has to be sent to itself by an automorphism. Clearly an automorphism has to send $\beta_{k} : x \mapsto x^{k}$ (and $y \mapsto y$), for $\gcd(n, k) = 1$, so we get the $(\Z/n\Z)^{\times}$ part.

Now $y$ can be sent by an automorphism to any involution $y x^{i} \in D_{2n} \setminus \Span{x}$. Assume this for a moment (will write it up later), and write $$ \alpha_{i} : y \mapsto y x^{i}, x \mapsto x. $$ Then one notes easily that $$ \alpha_{i} = \alpha_{1}^{i}, $$ so that $\Set{ \alpha_{i} : i \in \Z } = \Span{ \alpha_{1} }$ is a cyclic group of order $n$, hence isomorphic to $\Z / n \Z$.

And finally compute the conjugate $$ \beta_{k}^{-1} \alpha_{1} \beta_{k} $$ as $$ \beta_{k}^{-1} \alpha_{1} \beta_{k}(x) = \beta_{k}^{-1} \alpha_{1} (x^{k}) = \beta_{k^{-1}}(x^{k}) = x, $$ and $$ \beta_{k}^{-1} \alpha_{1} \beta_{k}(y) = \beta_{k}^{-1} \alpha_{1} (y) = \beta_{k^{-1}}(y x) = y x^{k^{-1}} = \alpha_{k^{-1}} $$ to see that $\Span{ \alpha_{1} }$ is normal in the automorphism group.