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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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    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 $h^\phi$.2012-06-21

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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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    No, I don't know any special name for that semidirect product, but someone may have thought of one.2012-06-22
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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 $G$).

The notion of autocommutator subgroup was introduced by Hegarty in The absolute center of a group, J. Algebra 169 (1994), 929-935.

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    ${}$(Just sayin')2012-06-21