Question about $\operatorname{Aut}(D_\infty)\cong D_\infty$

Question

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 $\omega$?

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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 $\psi_b$.

Answer 1

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.

Answer 2

Note that $\langle ab \rangle$ is cyclic group of infinite order and is subgroup of index 2 in $D_\infty$.

$D_\infty=\langle ab \rangle \sqcup b\langle ab \rangle$.

Note that elements in $\langle ab \rangle$ have infinite order, elements in $b\langle ab \rangle$ have order 2.

$\psi\in\operatorname{Aut}(D_\infty)$ preserves order of elements. Suppose $\psi(ab)=(ab)^p, ab=\psi((ab)^q)$ for $p,q \in \mathbb Z$.

Then $(ab)=(ab)^{pq}$, $p=q=1$ or $p=q=-1$.

$1$. If $\psi(ab)=ab$, then

$\psi$ must have form $\psi_{1,m}(a)=(ba)^m\cdot a$, $\psi_{1,m}(b)=(ba)^m\cdot b$ for some $m\in \mathbb Z$.

$\psi_{1,m}(a\cdot(ba)^m)=a,\ \psi_{1,m}(b\cdot(ba)^m)=b$, so $\psi_{1,m}$ is indeed an automorphism of $D_\infty$.

Define $\sigma,\in \operatorname{Aut}(D_\infty)$ by $\sigma(x)=axa^{-1}=axa$, so $\sigma(a)=a, \sigma(b)=aba,\ \sigma^2=\text{id}$.

Define $\omega\in \operatorname{Aut}(D_\infty)$ by $\omega(a)=b,\omega(b)=a$, so $\omega^2=\text{id}$. Then we have $\psi_{1,m}=(\omega\circ\sigma)^m$.

$2$. If $\psi(ab)=ba$, then

$\psi$ must have form $\psi_{2,m}(a)=(ba)^m\cdot b$, $\psi_{2,m}(b)=(ba)^m\cdot a$ for some $m\in \mathbb Z$.

$\psi_{2,n}(b\cdot(ba)^n)=a,\ \psi_{2,n}(a\cdot(ba)^n)=b$, so $\psi_{2,n}$ is indeed an automorphism of $D_\infty$,

and $\psi_{2,n}=\omega\circ(\sigma\circ\omega)^n$.

Combining 1 and 2, we have $\operatorname{Aut}(D_\infty) =\{\psi_{1,m}, \psi_{2,n}|m,n\in \mathbb Z\}$ is generated by $\sigma$ and $\omega$

with $\sigma^2=\omega^2=\text{id}$.

Thus $\operatorname{Aut}(D_\infty)=\langle \sigma, \omega|\sigma^2=\omega^2=\text{id}\rangle \cong D_\infty$. $\quad\Box$

Procedures in bold face are left for check.$\quad$ :)