I am trying to calculate $\text{Aut} (D_3)$, the automorphism group of the group of symmetries of the triangle. But I got stuck and now I have two questions about this.
Let me share my thoughts first:
First I noted that an isomorphism maps elements to elements of same order. Therefore, any $\varphi : D_3 \to D_3$ maps reflections to reflections and rotations to rotations. As candidates I got:
(1) the identity map $\varphi_1 (x) = x$
(2) the map $\varphi_2 (x) = x$ except $R_{120}\mapsto R_{240}$ and $R_{240}\mapsto R_{120}$
(3) the map $\varphi_3 (x) = x$ except $F_1 \mapsto F_2$ and $F_2 \mapsto F_1$
(4) the map $\varphi_4 (x) = x$ except $F_2 \mapsto F_3$ and $F_3 \mapsto F_2$
(5) the map $\varphi_5 (x) = x$ except $F_1 \mapsto F_3$ and $F_3 \mapsto F_1$
Then since $F_1 F_2 = R_{120}$, $\varphi_3 (R_{120}) = R_{240}$ and $\varphi_3 (R_{240}) = R_{120}$. Now I wanted to show that also $\varphi_2 (F_1) = F_2 $ and $\varphi_2(F_2) = F_1$ but unfortunately this doesn't have to be true since $\varphi_2$ could map $F_1$ to $F_3$ for example.
Now my question is:
Can one say something like we relabel the reflections and therefore
$\varphi_2 (F_1) = F_2$?
If yes then all these are equal and there are only two automorphisms —
That I have found anyway because my next question is:
How can I be sure that I found all automorphisms of a given group? Are
there upper and lower bounds on the number?
Best Answer
Let's characterize $D_3$ as the group generated by $r,t$ with $$r^3 = t^2 = (rt)^2 = e$$ $\hspace{10pt}$Suppose that $\phi:D_3 \to D_3$ is an automorphism.
Then, $\phi(t)$ must be of order two $\Rightarrow$ $\phi(t) \in \{t,rt, r^2 t\}$.
Similarly we can see that $\phi(r)$ must be of order three $\Rightarrow$ $\phi(r) \in \{r,r^2\}$.
Since these two choices determine $\phi$ (and any two are compatible), we have $2 \times 3 = 6$ possible choices for $\phi$.