Finding $\text{Aut}(D_3)$

Question

The following is a problem from my algebra homework:

Find all automorphisms of $D_3$ and determine which group $\text{Aut}(D_3)$ is isomorphic to.

I am fairly new at abstract algebra so this problem is somewhat of a challenge. I understand that automorphisms are isomorphisms from $G \to G$. However, I am not aware of a efficient/systematic approach to this problem since $D_3$ is not abelian. So far I have tried the brute force method: coming up with isomorphisms from $D_3 \to D_3$, but this seems time-consuming and not the most efficient way. Is there a way of telling how many elements there are in $\text{Aut}(D_3)$ (or better yet, for a general group $G$) so that I know when I am done? What is an efficient approach to this problem?

Answer 1

Let's characterize $D_3$ as the group generated by $\sigma,\tau$ with $\sigma^3 = \tau^2 = (\sigma \tau)^2 = e$.

Suppose that $\phi:D_3 \to D_3$ is an automorphism.

$\phi(\tau)$ must be of order $2$. So, we must have $\phi(\tau) \in \{\tau,\sigma \tau, \sigma^2 \tau\}$.

$\phi(\sigma)$ must be of order $3$. So, we must have $\phi(\sigma) \in \{\sigma,\sigma^2\}$.

Since these two choices determine $\phi$ (and any two are compatible), we have $2 \times 3 = 6$ possible choices for $\phi$.

In general, it is usually most effective to look at how a homomorphism acts on some set of generators.

Answer 2

There is a (relatively) easy way to find automorphisms of $D_3$. It is known that $D_3$ has following presentation:

$$\langle r,d\mid r^3=d^2=(rd)^2=e\rangle$$

That is, $D_3$ is generated by elements $r$ and $d$ satisfying $r^3=d^2=(rd)^2=e$. Also, you can check that $e,r,r^2,d,rd,r^2d$ are only elements of that group.

If $\varphi:D_3\to D_3$ is an automorphism, then $\varphi(r)^3=e$, $\varphi(d)^2=e$ and $(\varphi(r)\varphi(d))^2=e$. In addition, every homomorphism $\psi:D_3\to G$ is determined by the value of $\psi(r)$ and $\psi(d)$. (Since $\{r,d\}$ generates $D_3$.)

Since the order of $r$ and $r^2$ is 3 and the order of $d$, $rd$ and $r^2d$ is 2, possible values of $\varphi(r)$ is $r$ and $r^2$, and possible values of $\varphi(d)$ is $d$, $rd$ and $r^2d$. You can check that each cases defines an automorphism of $D_3$.

Answer 3

Another presentation for $D_3 = S_3$ is $D_3 = \langle{\sigma, \tau:\, \sigma^3 = 1, \tau^2 = 1, \tau^{-1}\sigma\tau = \sigma^{-1}\rangle}$. It's then easy to show that the center of $D_3$ is thus $1$, and hence $\text{Aut}(D_3)$ contains the subgroup $\text{Inn}(S_3) = S_3$. On the other hand, any $f\in \text{Aut}(D_3)$ is determined by its values on $\sigma, \tau$, and there are three nontrivial elements of order $2$ and two nontrivial elements of order $3$ in $D_3$. It follows the map $f \to (f(\sigma), f(\tau))$ is injective and has order at most $6$. Hence the inclusion $\text{Inn}(D_3) \to \text{Aut}(D_3) = S_3$ is an isomorphism.

In the general setting, there isn't a nice way of determining $\text{Aut}(G)$ for a group $G$. You can certainly consider all bijections $f:G \to G$ and restrict to ones that preserve order, but that's not a useful algorithm for larger groups.