# Automorphisms “killing” and group

I ran into the following concept in passing here. Let $$G$$ be a group and let $$\phi$$ be an automorphism of $$G$$. Let $$P$$ be a presentation for $$G$$ with $$X$$ the set of generators in $$P$$. Form a new group $$P^\phi$$ by appending to $$P$$ relations of the form $$x = \phi(x)$$ for all $$x \in X$$.

Is the isomorphism type of $$P^\phi$$ independent of the choice of $$P$$? Maybe I should assume $$G$$ is finitely presented and we are just working with finite presentations.

We say that $$\phi$$ "kills" $$G$$ (I know a bit dramatic) if $$P^\phi = 1$$ for some (all?) choices of $$P$$. Is there an equivalent way of saying that $$\phi$$ kills $$G$$ that doesn't mention presentations?

• As to your first question, yes, $P^{\phi}$ is independent of the choice of $P$, as $x = \phi(x)$ for all $x \in X$ is equivalent to $g = \phi(g)$ for all $g \in G$. – Andreas Caranti Apr 24 at 7:09
• This is usually called a semidirect product of $G$ with $Z$. – Moishe Kohan Apr 24 at 15:31

Since you like topology, let $$M$$ be a space with $$\pi_1(M)=G$$ and let $$\phi$$ be induced by a map $$f:M\to M$$. If you construct the mapping torus $$M_f=M\times [0,1]/(x,1)\sim(f(x),0),$$ its fundamental group is given by an HNN extension $$\pi_1(M_f) = \langle G,t\mid tgt^{-1}=\phi(g)\text{ for all g\in G}\rangle.$$ The $$t$$ corresponds to a loop that closes up the path $${*}\times [0,1]$$ with some arc in $$M$$ (or: choose $$f$$ so that the basepoint of $$M$$ is a fixed point). If you then glue in a disk $$D$$ along this loop, you get $$\pi_1(M_f\cup D)=\langle G \mid g=\phi(g)\text{ for all g\in G}\rangle.$$ That this is independent of $$M$$, $$f$$, and $$D$$ is that the group presentation on the right does not mention them. Also, notice that if $$G$$ is finitely presented, so is $$\pi_1(M_f\cup D)$$ since using the presentation complex for $$M$$ gives $$M_f\cup D$$ a finite CW structure. (Equivalently, one can take $$\pi_1(M_f/\partial D)$$ to get the group in question.)
Underlying this is the normal closure of the subgroup $$\langle g^{-1}\phi(g):g\in G\rangle$$. The map $$\phi$$ "kills" $$G$$ iff the normal closure is all of $$G$$.
A weird equivalent statement is that $$\phi$$ "kills" $$G$$ iff, for every normal covering space of $$M_f$$ that $$\partial D$$ lifts to, that covering space has only one sheet.
Something I couldn't find a use for is that if $$M$$ is a $$K(G,1)$$, then $$M_f$$ is a $$K(\pi_1(M_f),1)$$. The problem is that collapsing the $$*\times [0,1]$$ loop seems like it might create nontrivial higher homotopy groups even if $$\phi$$ "kills" $$G$$. It would be interesting if $$\phi$$ "killing" $$G$$ means the collapsed $$K(\pi_1(M_f),1)$$ is contractible.
As an example, for a fibered knot $$K\subset S^3$$, the complement $$S^3-\nu(K)$$ is a mapping torus, with the monodromy inducing an automorphism $$\phi:\pi_1(S)\to \pi_1(S)$$ where $$S$$ is some leaf, a once-punctured compact oriented surface. The monodromy is isotopic to one with a fixed point on the boundary of $$S$$, so there is a meridian that serves the role of $$t$$ for $$\pi_1(S^3-\nu(K))$$ as an HNN extension. Gluing in a thickened disk along this meridian results in a space that is homeomorphic to $$S^3-B^3$$, so not only does $$\phi$$ "kill" $$\pi_1(S)$$, but the resulting collapsed mapping torus is contractible.