# What happens when you internalize outer automorphisms?

Given a finitely presented group $G = (Gen|Rel)$, we have a set of inner automorphisms $\{ \phi_a(x) = axa^{-1} | a \in G\}$. Defining the set of outer automorphisms to be those automorphisms of $G$ which are not in the inner set, given an outer automorphism $\phi(x)$, we can create a new group which has the presentation $(Gen \cup \{\psi\}|Rel \cup \{ \psi x \psi^{-1} = \phi(x) | x \in Gen\})$. I've been referring to this notion by saying we want to internalize the outer automorphism $\phi(x)$. Some questions I've had so far are as follows:

What properties of the original group are preserved by this process?

Do most groups become trivial or infinite? (So far its looking like the latter)

I find this to be really difficult so I really hope somebody here will have some answers for these questions, or at least a place to look or a reason to expect this to be very difficult to answer. So far I've done this for some cyclic groups and they become evidently massive, though I can't prove that because I don't know how, as I just started reading about finite presentations and this sort of thing recently.

• I believe the construction you describe is known as an HNN extension (for Higman Neumann, and Neumann). – Andreas Blass Mar 1 '17 at 1:39
• This is exactly a semidirect product $G \rtimes \mathbb{Z}$ (so it has the same cardinality as $G \times \mathbb{Z}$). HNN extensions are, as you say, a bit more general. I don't know what kind of properties you have in mind. – Qiaochu Yuan Mar 1 '17 at 2:14
• If your outer automorphism $\phi$ has finite order $n$ (as an automorphism), then you could impose the additional relation $\psi^n=1$ on your new group to get a semidirect product $G \rtimes C_n$. So this would preserve finiteness if $G$ was finite. – Derek Holt Mar 1 '17 at 3:01
• Isn't this just the subgroup of the holomorph generated by $G$ and $\psi$? Given a group $G$, since $\mathrm{Aut}(G)$ acts on $G$, you can define a semidirect product $G\rtimes \mathrm{Aut}(G)$; your group is the subgroup generated by $G$ and $\psi$. – Arturo Magidin Mar 1 '17 at 4:23
• @ArturoMagidin: This would only be true if you would add the relation $\psi^n=1$ (where $n$ is the order of $\phi$) as indicated by Derek's comment. – Tom De Medts Mar 1 '17 at 9:40

Your group is just the semidirect product $\Gamma=G\rtimes_\psi\mathbb{Z}$. It's easy to check that the isomorphism class of $\Gamma$ only depends on the conjugacy class of $\psi$ in $\mathrm{Out}(G)$; in particular, if $\psi$ were inner this would be the direct product.
One trivial corollary is that such $\Gamma$ is always infinite. (You could also have seen this using presentations by spotting the homomorphism to $\mathbb{Z}$ that kills $G$.)
• Does your statement about the isomorphism class depending only on the conjugacy class of $\psi$ require that $G$ does not surject onto $\mathbb{Z}$? – ADL Apr 7 '17 at 13:03
• @ADL -- I only meant that if one replaces $\psi$ by a conjugate, it doesn't change the isomorphism class of $\Gamma$. It is of course also true that a given $\Gamma$ can arise from different conjugacy classes $[\psi]$ when the abelianization of $\Gamma$ has rank greater than 1. – HJRW Apr 7 '17 at 13:06
• Ah, sorry, I over-interpreted your statement. If $G$ does not map onto $\mathbb{Z}$ then these statements are essentially equivalent, which is what I thought you were saying (that is, if $G$ does not surject onto $\mathbb{Z}$ then $G\rtimes_{\psi_1}\mathbb{Z}\cong G\rtimes_{\psi_2}\mathbb{Z}$ if and only if $[\psi_1]$ is conjugate to $[\psi_2]$ or to $[\psi_2^{-1}]$ modulo the inner automorphisms (by $[\psi]$ I mean the outer automorphism $\psi\operatorname{Inn}(G)$).) – ADL Apr 7 '17 at 15:14