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The notion "holomorph" was introduced in Maria S. Voloshina's Ph.D. thesis On the Holomorph of a Discrete Group (available at https://arxiv.org/abs/math/0302120). It is defined as follows:

Let $G$ be a group and let $\mathrm{Aut}(G)$ be the automorphism group of $G$. The holomorph of $\mathrm{Hol}(G)$, is defined as follows:

  1. As a set, $\mathrm{Hol}(G)=\mathrm{Aut}(G)\times G$;
  2. For each $x,y\in G$ and $f,g\in\mathrm{Aut}(G)$, the multiplication on $\mathrm{Hol}(G)$ is defined by $$(f,x)\cdot(g,y)=(fg, g^{-1}(x)y)\text{.}$$

The author points out that there is a split exact sequence $$1\to G\to\mathrm{Hol}(G)\stackrel{\leftarrow}{\rightarrow}\mathrm{Aut}(G)\to1\text{.}$$

What I am confused about is, if there is such a split exact sequence, doesn't the splitting lemma imply that $\mathrm{Hol}(G)$ is isomorphic to $G\times\mathrm{Aut}(G)$?

Alp Uzman
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zuriel
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  • Fix your sequence. There is no $K$. – KCd Jan 12 '15 at 17:23
  • @KCd, thanks! It was fixed. – zuriel Jan 12 '15 at 17:24
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    Your groups are not generally abelian, so that "splitting lemma" does not apply. The splitting you describe corresponds to an isomorphism with a semidirect product, and indeed the group structure you describe on the holomorph makes it a semidirect product of $G$ and ${\rm Aut}(G)$. See Theorems 3.2 and 3.3 of http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/splittinggp.pdf. – KCd Jan 12 '15 at 17:28
  • Thanks @KCd! I forgot the condition on the splitting lemma and now it is very clear to me. Am wondering if you happen to know anything about my latest question at http://mathoverflow.net/questions/193755/very-frustrated-reading-a-proof-of-the-faithfulness-of-artins-representation-of – zuriel Jan 12 '15 at 17:41
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    While Voloshina's thesis (2003) is about holomorphs, that concept was not introduced there for the first time. I can find it, with that name, in Birkhoff and Mac Lane's Algebra from the 1960s and the Wikipedia page on holomorphs lists Marshall Hall's group theory book from 1959 at the end. – KCd Jan 12 '15 at 17:41
  • Thanks for introducing the history of holomorph, @KCd!! – zuriel Jan 12 '15 at 17:48

1 Answers1

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This is an answer to the question that seems to have already been resolved in the comments.

As Prof. Conrad noted, the holomorph $\text{Hol}(G)$ of a group $G$ is isomorphic to the semidirect product $G\rtimes_\sigma \text{Aut}_{\text{Grp}}(G)$, where $\text{Aut}_{\text{Grp}}(G)$ is the group of group automorphisms of $G$, and $\sigma=\text{id}_{\text{Aut}_{\text{Grp}}(G)}:\text{Aut}_{\text{Grp}}(G)\to \text{Aut}_{\text{Grp}}(G)$ is the standard action.

An example of this group (interpreted at least measurably, so that the automorphisms are required to be at least Borel measurable; see e.g. Measurable group homomorphisms are continuous) when $G$ is commutative is the so-called affine group; one can similarly consider affine transformations of tori; see Example of a continuous affine group action, Question about Stillwell Naive Lie theory exercise 4.6.3, Dynamics on the torus. The second discussion is especially relevant. Indeed, note that $\text{Hol}(\mathbb{R})$ is isomorphic to

$$\mathbb{R}\rtimes \text{GL}(1,\mathbb{R})\cong (\mathbb{R},+)\rtimes (\mathbb{R}\setminus\{0\},\cdot), $$

so that both groups involved are abelian, and there is a short exact sequence

$$0\to\mathbb{R}\to \text{Hol}(\mathbb{R})\to \text{GL}(1,\mathbb{R})\to 1,$$

however this sequence is not in the category of abelian groups (it's in the category of possibly nonabelian groups).

Finally as a historical comment, the notion of the "holomorph of a group" goes at least to Burnside's 1897 book Theory of Groups of Finite Order (available at https://archive.org/details/cu31924086163726).

Alp Uzman
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