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

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 $$\psi_b$$.

• It should be noted that $\Bbb Z_2 * \Bbb Z_2$ here means the free product of the two groups. – Cameron Williams Nov 4 '19 at 13:40
• You should use the fact that the subgroup $\langle ab \rangle$ is infinite cyclic and is characteristic and of index $2$ in $D_\infty$ – Derek Holt Nov 4 '19 at 13:52
• @CameronWilliams $a = b(ab)^{-1}$. – Unit Nov 4 '19 at 14:30
• @Unit Ah yep that's it. I only considered the positive powers. I think I need a second cup of tea this morning.. – Cameron Williams Nov 4 '19 at 14:48
• A modern-ish reference for automorphism groups of free products is: Gilbert, N. D. (1987), Presentations of the Automorphism Group of a Free Product. Proceedings of the London Mathematical Society, s3-54: 115-140. doi:10.1112/plms/s3-54.1.115 – user1729 Nov 4 '19 at 14:58

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.

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.$$\quad$$ :)