What is the automorphism group of $D_5$ (and, more generally, of $D_n$)?

Question

Here is my solution:

Recall how $D_5$ is defined. We have 4 non-identity rotations:

$$ T, T^2, T^3, T^4, $$

with $T^0 = 1$. We have the reflection $S$, such that $S^2 = 1$. To connect reflected rotations to rotations, we have: $STS = T^{-1}$. These relationships define $D_5$ in their entirety.

Let $D'_5$ be ismorphic to $D_5$, with rotations: $$ 1, U, U^2, U^3, U^4,$$

$U^0 = 1$, the reflection $V$, such that $V^2 = 1$, and the relationship $VUV = U^{-1}$.

Thus, if one were creating an isomorphism $f: D_5 \mapsto D'5$, we are only free to make one choice: we can map some rotation (reflected or otherwise) to $U$, in other words, let $U = S^aT^j$, for some $a \in \{0, 1\}$ and some $j \in \{1, 2, 3, 4\}$. Once this choice is made, all the other rotations in $D'_5$ are determined:

$$ 1, U = S^a T^j, U^2 = UU, \text{etc.} $$

We must have that $S = V$, since there is no other option. Since we have the rule $VUV = U^{-1}$, so all the $VU^kV$ for $k \in \{0, 1, 2, 3, 4\}$ are also determined.

So, we have 8 choices to make for our one option: we can let $U$ be one of the $4$ rotations in $\{k \in \{1, 2, 3, 4\} \mid T^k\}$ or one of the 4 reflected rotations: $\{k \in \{-1, -2, -3, -4\} \mid T^k \}$, and thus there are 8 automorphisms.

Is this correct?

Answer 1

$D_5$ can be presented as a group generated by two elements of order 2 whose product has order 5.* Any pair of reflections in $D_5$ have this property. Thus, a fixed such pair can be mapped to any other such pair uniquely determining all the automorphisms. There are 5 reflections and so 20 (5 x 4) such ordered pairs. Thus $D_5$ has 20 automorphisms. 10 are inner automorphisms and arise from symmetries of the plane. The others do not so arise as they map reflections separated by 36 degrees to reflections separated by 72 degrees.

In fact, by the same reasoning, for any odd prime $p$, $D_p$ has $p(p-1)$ automorphisms, given by the elements of its normalizer in $S_p$ (when viewed as acting on the vertices of a $p$-gon).

Indeed, for any $n\ge 3$, the number of reflections in $D_n$ that pair with a fixed reflection such that their product has order $n$ is $\varphi (n)$, the number of integers less than $n$ and coprime to $n$. This is because the product of a fixed reflection by a variable reflection gives each of the possible rotations once, and $\varphi (n)$ of them have order $n$. This, for $n\ge 3$, $|Aut(D_n)|=n\varphi (n)$.

For $n$ odd, we can say more. In this case, $D_n$ is centerless and since if we take the natural embedding in $S_n$ given by the action on the vertices of an $n$-gon we have $|N_{S_n}(D_n)|=n\varphi(n)$, the automorphisms are all given by the normalizer in $S_n$, which also happens to be the normalizer of an $n$-cycle in $S_n$. Thus $Aut(D_n)$ can be thought of as all the permutations of $\Bbb{Z}/(n)$ given by $i\mapsto ai+b$ for $b\in\Bbb{Z}/(n)$ and $a\in\Bbb{Z}/(n)^{\times}$.

*To see that $G=<a,b|a^2=b^2=(ab)^5>$ is isomorphic to $D_5$ note that any two reflections in $D_5$ satisfy all those relations, so $D_5$ is a homomorphic image of $G$. Now, consider the order of $G$. Any word with consecutive a's or b's can be reduced, as can any word containing $ababababab$. This leaves at most the empty word, 9 words starting with $a$, and 10 words starting with $b$. But also $ababababa=b$, $abababab=ba$, etc., so there are at most 10 distinct elements of $G$. (Actually, you don't need the etc.; even without it you have that there are fewer than 20 distinct elements, and you already know $|G|$ is divisible by 10 since $D_5$ is an image.) Thus $G$ is isomorphic to $D_5$.