TITLE: Why are automorphisms of $D_{2n}, n \geq 5$ odd, not always inner?
QUESTION [5 upvotes]: The dihedral group of order $2n$ is often presented as $$D_{2n}= \langle r,s: r^n=s^2=1, rs= sr^{-1} \rangle \text{,}$$ where $r$ denote rotations and $s$ denote reflections of a regular $n$-gon. Furthermore, I've read that for $D_6 \cong S_3$, i.e. the symmetric group of permutations on $3$ letters, every automorphism is inner. I would like to know why this is so for this particular case, and why then is not every automorphism inner for $D_{2n}, n \geq 5$, where $n$ is odd? For the first case, $Z(D_6)=1$, i.e. the center of $D_6$ is trivial, so that the group of inner automorphisms $\text{Inn}(D_6) \cong D_6$, but how can I justify that such automorphisms are actually inner?

REPLY [2 votes]: With the risk of repeating what was said before...
$D_{2n}$ is the group of symmetries of a regular polygon with $n$ sides (and is sometimes called $D_n$... but for us $D_{2n}$ has $2n$ elements).
In the presentation of $D_{2n}$ with $r$ and $s$, denote $t:=s r$. Then the equality $rs = sr^{-1}$ is equivalent to (using also $s^2 = 1$) $sr sr = 1$, that is, $t^2=1$. Moreover, $st=r$ satisfies $(st)^n=1$. Hence, an equivalent presentation of $D_{2n}$ is $\langle s,t \ \colon \ s^2 = t^2=1, (st)^n=1\rangle$. 
To obtain the presentation of $D_{2n}$  from above we can take $s$, $t$ to be reflections of the regular polygon with respect to two axes that form an angle of $\frac{\pi}{n}$. Then $st$ is a rotation of angle $\frac{2\pi}{n}$. The order of $s$, $t$ and $st$ are respectively, exactly $2,2,n$.
A bit more about the presentation: if we have a group $D$ generated by two elements $s$, $t$ of order $2$ and $N$ is the order of $st$ then $D$ is naturally isomorphic to $D_{2N}$. 
Assume  now on that $n$ is odd. There are $n$ elements of order $2$ in $D_{2n}$, the $n$ reflections with respect to axes passing through a vertex of the polygon and its center. Any two such reflections are conjugate elements, since any such axis can be taken to another axis by a transformation of the polygon. 
Here is how to produce all the automorphisms of $D_{2n}$. We need to find the images $s_1$ and $t_1$ of $s$ and $t$. These have to be elements of order $2$ and the order of $s_1 t_1$ is $n$. 
Assume  now on that $n$ is odd. There are $n$ elements of order $2$ in $D_{2n}$, the $n$ reflections with respect to axes passing through a vertex of the polygon and its center. Any two such reflections are conjugate elements, since any such axis can be taken to another axis by a transformation of the polygon. 
Then $s_1$, $t_1$ are also reflections with respect to two axes of symmetry of the polygon.  Since the order of $s_1t_1$ is $n$, the angle between these axes is $\frac{k \pi}{n}$ where $\gcd(k,n)=1$.  
The automorphism $s\mapsto s_1$, $t\mapsto t_1$ is a conjugation by $\phi$ if and only if the tranformation $\phi$ of the polygon takes the axis of $s$ to the axis of $s_1$ and the axis of $t$ to the axis of $t_1$. But  such a $\phi$ exists if and only if the angles between these pairs of axes are the same. This may not be always be true  for odd $n>3$. ( and also for even $n$ > 4..). 
Here is a picture for the group of a regular pentagon.

The red reflections map to blue ones. This gives an automorphism of $D_{10}$ which is not inner.