# Is there a name of semidirect product of a group with its automorphism group?

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?

• I think it is sometimes called the holomorph (of $G$) in early texts. – Geoff Robinson Oct 31 '18 at 15:06
• 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]. – S. Carnahan Oct 31 '18 at 15:16
• See this Wikipedia article. – abx Oct 31 '18 at 15:17
• @GeoffRobinson - to my knowledge, it's still called the holomorph. I first met this word in Joseph Rotman's book. – benblumsmith Oct 31 '18 at 15:32
• @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. – Geoff Robinson Oct 31 '18 at 15:37

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 $$\operatorname{Aut}(G)$$.
• 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 $G$. – Breakfastisready Nov 1 '18 at 12:53