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?

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    $\begingroup$ I think it is sometimes called the holomorph (of $G$) in early texts. $\endgroup$ Oct 31, 2018 at 15:06
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    $\begingroup$ 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, 2018 at 15:16
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    $\begingroup$ See this Wikipedia article. $\endgroup$
    – abx
    Oct 31, 2018 at 15:17
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    $\begingroup$ @GeoffRobinson - to my knowledge, it's still called the holomorph. I first met this word in Joseph Rotman's book. $\endgroup$ Oct 31, 2018 at 15:32
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    $\begingroup$ @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. $\endgroup$ Oct 31, 2018 at 15:37

1 Answer 1


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 $\operatorname{Aut}(G)$.

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    $\begingroup$ 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 $G$. $\endgroup$ Nov 1, 2018 at 12:53

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