Infinite dihedral group $D_\infty:=\langle a,b|\ a^2=b^2=1\rangle = \mathbb Z_2 * \mathbb Z_2$.
For $x\in D_\infty$, define $\psi_a, \psi_b\in\operatorname{Inn}(D_\infty)$ by $\psi_a(x)=axa^{-1}, \psi_b(x)=bxb^{-1}$.
Define $\omega\in \operatorname{Aut}(D_\infty)$ by $\omega(a)=b,\ \omega(b)=a$.
$\psi_a^2=\psi_b^2=\omega^2=1,\psi_a\omega=\omega\psi_b$. This can be reduced to $\psi_a^2=\omega^2=1$.
If we can show $\operatorname{Aut}(D_\infty)$ is generated by $\psi_a,\psi_b$ and $\omega$, then $\operatorname{Aut}(D_\infty)\cong D_\infty$.
My question:
How can we prove $\operatorname{Aut}(D_\infty)$ is generated by $\psi_a,\psi_b$ and $\omega$?
Update:
Thanks to Derek Holt and Unit, I gave an answer below, using nearly the same notation, except for $\psi_a$ being relaced by $\sigma$ and without using $\psi_b$.
Best Answer
Let's use the more intuitive presentation $$D = D_{\infty} = \langle r, f | f^2 = 1, \ frf = r^{-1} \rangle$$ which is equivalent to yours by $f \mapsto a$ and $r \mapsto ab$.
First note that $C = \langle r \rangle$ is an infinite cyclic subgroup of $D$ of index 2, hence normal. Better, it's characteristic: it comprises precisely all the elements of $D$ of infinite order, and since automorphisms preserve order, they cannot send elements of $C$ outside of $C$. Thus every automorphism of $D$ restricts to an automorphism of $C$, so if $\rho \in \operatorname{Aut}(D)$ then $\rho(r) = r^{\pm 1}$.
Next, $\rho(f) = fr^k$ for some integer $k$ without restriction, because $\rho$ must fix the coset $fC$ as a set. Thus the automorphisms are parametrized by pairs of signs and integers. You should check that such pairs indeed give automorphisms of $D$ and that they satisy the obvious composition law that makes their aggregate a dihedral group.