# What is this automorphism-related subgroup?

Let $G$ be a group, let $H\leq G$ and let $\phi\in\operatorname{Aut}(G)$. Then what "is" the subgroup $K=\langle h^{-1}(h\phi): h\in H\rangle$?

Does it, or its normal closure, have a name? Does it have any interesting properties?

Stuff seems to get interesting if you assume $H\phi=H$, but what about in the general case?

(Really, I just want to know if it has a name so I can search for stuff about it - but any information would be nice. I'm not fussy!)

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(You're writing automorphisms as right actions I take it?) –  anon Jun 21 '12 at 15:05
Doesn't everyone?! –  user1729 Jun 21 '12 at 15:08
@user1729: No, we don't. –  Arturo Magidin Jun 21 '12 at 15:09
@ArturoMagidin: Pity. –  user1729 Jun 21 '12 at 15:10
Chapter 8 "Groups acting on Groups" in Kurzweil–Stellmacher's book, and chapters 2 and 5 in Gorenstein's Finite Groups are probably good. The key thing is to think of $H \leq G \leq \operatorname{Hol}(G) = \operatorname{Aut}(G) \ltimes G$ and to write right actions as conjugation $h^\phi$. –  Jack Schmidt Jun 21 '12 at 17:18

The group you are talking about is $[H, \phi].$ This group, though it does not really have a special name, is well known to group-theorists and goes back much further than the reference given by Arturo. More generally, if $A$ is any group of automorphisms of the group $G,$ then the group $[G,A]$ is the group generate by $\{ g^{-1}g^{a}: g \in G, a \in A \}.$ Here the result of the automorphism $a$ on $g$ is denoted by $g^{a},$ which is consistent with thinking of $[G,A]$ as a subgroup of the semidirect product $GA.$ Denotin, as is standard, $g^{-1}g^{a}$ by $[g,a]$ is is easy to check that $[g,a]^{h}[h,a] = [gh,a]$ for all $a \in A$ and $g,h \in G$ while also $[g,ab] = [g,b][g,a]^{b}$ for all $g \in G$ and $a,b \in A.$ This shows that for any $a \in A$, $[G,a]$ is a normal subgroup of $G$ ( so $[G,A] \lhd G$ also), and $[G,A]$ is $A$-invariant. This (sketchy) discussion is all at least implicit in D. Gorenstein's 1968 book "Finite Groups", for example. Since $H$ is not $\phi$-invariant in your question, we need to be a little more careful. Since $[xy,\phi] = [x,\phi]^{y}[y,\phi]$ for $x,y \in H,$ it does follow that $[H.\phi]$ is normalized by $H$, though it need not be a subgroup of $H.$ Similarly, the equation $[h, \phi^{2}] = [h,\phi][h,\phi]^{\phi}$ implies that $[H, \langle \phi \rangle ]$ is a $\phi$-invariant subgroup. For the moment, I am unsure whether $[H, \phi] = [H, \langle \phi \rangle]$ in general, though this is true when $H$ is $\phi$-invariant-(added later)-indeed, as Jack Schmidt points out, $[H,\phi] \neq [H, \langle \phi \rangle ]$ in general.

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Why is it $H$-invariant? Surely you would need $H\phi=H$ for this to happen? –  user1729 Jun 21 '12 at 15:52
I didn't read question carefully enough. Thought $\phi$ was an automorpism of $H$, didn't take in that it was an automorphism of a bigger group $G$ and that $H$ need not be $\phi$-invariant. Will re-edit. –  Geoff Robinson Jun 21 '12 at 16:25
Let $\phi$ have order 3, $G$ be Klein 4, $H$ be order 2. Then $[H,\phi]$ has order 2 and is not $\phi$-invariant, but $[H,\langle\phi\rangle]=G$ has order 4 and is $\phi$-invariant. –  Jack Schmidt Jun 21 '12 at 17:16
@Jack: nice example –  Geoff Robinson Jun 21 '12 at 17:47
Do you know if there is a special name for $G\rtimes A$? I know that if $A=Aut(G)$ then this is the holomorph...would you call it the $A$-holomorph, say? –  user1729 Jun 22 '12 at 10:54
It is related to the autocommutator subgroup of a group. If $G$ is a group, the autocommutator subgroup of $G$ is generated by all elements of the form $g\phi(g)^{-1}$, where $\phi$ ranges over all automorphisms of $G$, and $g$ over all elements of $G$.
Restricting the collection of automorphisms to some subgroup $M$ of $\mathrm{Aut}(G)$ gives the $M$-commutator subgroup. For example, if $M=\mathrm{Inn}(G)$, then the $M$-commutator subgbroup of $G$ is just the usual commutator subgroup of $G$.
You are looking at the restriction of the $\langle\phi\rangle$-commutator subgroup given by considering only elements of $H$, i.e., you are looking at $[H,\phi]$ (which can be seen as a subgroup of the holomorph of $G$).
${}$(Just sayin') –  user1729 Jun 21 '12 at 15:35