11
$\begingroup$

Consider the construction $G \rtimes \text{Aut}(G)$. Here $ G$ is a group, $\text{Aut}(G)$ is the automorphism group and the semidirect product is over the most obvious action.

1) Is there any name for such a general construction? To me, it seems like the most straight-forward example of a semi-direct product.

2) Are there any surveys over such constructions or any big theorems about the structure of such groups?

3) I'm specifically interested in finding torsional elements when $G = F_2$, the free group of two generators. Is there any result about this special scenario?

Edit: More thoughts about question 3 are below.

As pointed in the comments, if you have any torsional automorphism $\phi$ of the free group (there are good classification theorems for such automorphisms), then a torsional element of the semidirect product will be of the form $(g,\phi)$ such that $g\phi(g)\phi^2(g)... \phi^{k-1}(g)=1$, $k$ being the order of $\phi$.

You can check that any element of the form $(\alpha^{-1} \phi(\alpha),\phi)$ works where $\phi$ is a torsional automorphism and $\alpha \in F_2$. Are these all such elements? Is there a general form of such elements?

$\endgroup$
  • 25
    $\begingroup$ I think it is sometimes called the holomorph (of $G$) in early texts. $\endgroup$ – Geoff Robinson Oct 31 '18 at 15:06
  • 6
    $\begingroup$ One basic property is that the holomorph $\operatorname{Hol}(G)$ is the normalizer of $G$ in its permutation group under the left regular embedding. I've only seen it applied in Hopf Galois theory, where one has a bijection between Hopf Galois structures with group $G$ and "regular" embeddings of $G$ in its holomorph [Byott 1999]. $\endgroup$ – S. Carnahan Oct 31 '18 at 15:16
  • 9
    $\begingroup$ See this Wikipedia article. $\endgroup$ – abx Oct 31 '18 at 15:17
  • 4
    $\begingroup$ @GeoffRobinson - to my knowledge, it's still called the holomorph. I first met this word in Joseph Rotman's book. $\endgroup$ – benblumsmith Oct 31 '18 at 15:32
  • 3
    $\begingroup$ @benblumsmith :OK, thanks I think the word "holomorph" was used in W. Burnside's famous book (there were editions in 1898 and 1911). It seems that the nomenclature stuck. $\endgroup$ – Geoff Robinson Oct 31 '18 at 15:37
6
$\begingroup$

As a first remark, note that if $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ is a finite subgroup and $G$ is torsion-free, then the projection $p: G\rtimes \operatorname{Aut}(G)\to \operatorname{Aut}(G)$ maps $\tilde{H}$ isomorphically to some finite subgroup $H=p(\tilde{H})\leq \operatorname{Aut}(G)$.

Now for each finite subgroup $H\leq \operatorname{Aut}(G)$, you can ask yourself what are the subgroups $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ projecting to $H$? These correspond to splittings of the semi-direct product $G\rtimes H$, which correspond to $1$-cocycles (or crossed homomorphisms) $f:H\to G$, that is, functions satisfying $$ f(ab) = f(a){}^af(b),\quad a,b\in H. $$ Here ${}^ag$ denotes the action of $a$ on $g$. These $1$-cocycles represent elements of the first non-abelian cohomology set $H^1(H;G)$, the trivial element of which is represented by any cocycle of the form $f(a) = (g^{-1}){}^ag$ for some $g\in G$.

So in some sense the answer to your follow-up question lies in the non-abelian cohomolgy of finite subgroups of $\operatorname{Aut}(G)$.

$\endgroup$
  • 3
    $\begingroup$ Thanks. After I was trying to piece together an explanation like this. If we take $H$ to be the cyclic group generated by a single torsional element $\phi \in \text{Aut}(G)$ of order $k$ then any map $f \in H^1(H;G)$ must satisfy $1=f(1)=f(\phi^k) = f(\phi)\ ^{\phi}f(\phi)\ ^{\phi^2}f(\phi)...^{\phi^{k-1}}f(\phi) $. Also, any element of $f \in H^1(H;G)$ is known by the value it takes on $\phi$. So the question really comes down to understanding the non-abelian cohomology of $H$ over $G$. $\endgroup$ – Breakfastisready Nov 1 '18 at 12:53

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.