Recently, I have been studying complete groups, and I know that if a group $N$ is complete then for every short exact sequence $1 \rightarrow N \xrightarrow{f} G \xrightarrow{g} H \rightarrow 1$ of groups, we have $G = f(N)L$ for some $L \unlhd G$ such that $f(N) \cap L = \{e\}$. Does this characterize complete groups? Namely, I have the following question: if $N$ is a group such that for every short exact sequence $1 \rightarrow N \xrightarrow{f} G \xrightarrow{g} H \rightarrow 1$ of groups, we have $G = f(N)L$ for some $L \unlhd G$ such that $f(N) \cap L = \{e\}$, then does it imply that $N$ is a complete group? I am completely stuck with this question, please help me.
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2By embedding into the holomorph, you can see that your property on $N$ implies that $N$ has no outer automorphisms. I am not sure about trivial centre. In a counterexample, the centre would have to be a divisible abelian group, since otherwise you could extend $N$ nontrivially as a central product. – Derek Holt Jun 04 '24 at 07:51
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@Derek Holt, I understand that $Out(N) = 1$. Still, I could not understand the second paragraph you mentioned: "In a counterexample, the center would have to be a divisible abelian group since otherwise, you could extend N nontrivially as a central product", this part. It would be helpful if you could kindly clarify my doubts. – Pratina Jun 05 '24 at 04:39
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2Let $Z=Z(N)$. If $Z$ is not divisible then there is an abelian group $Y$ with $Z <Y$ such that $Z$ has no complement in $Y$. Then the central product $G$ of $N$ and $Y$ amalgamating $Z$ would not have your property because $Z$ would have no normal complement in $G$. – Derek Holt Jun 05 '24 at 07:23
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@Derek Holt, thank you very much, I understand it now clearly. I was wondering if we could get an example of a group $N$ such that $N$ is not complete and $N$ satisfies the above property. – Pratina Jun 05 '24 at 14:25
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You could try asking the question on mathoverflow, including a link to the question here. – Derek Holt Jun 05 '24 at 17:40
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The property you claimed is equivalent to a semi-direct product. What we do know is that the complete group property is equivalent to a direct product version, see theorem 7.17 of the book GTM148 Joseph J. Rotman (auth.) - An Introduction to the Theory of Groups-Springer-Verlag New York (1995).
Basic results about holomorph of a group (just above the theorem in this book), and the characterization of direct/semi-direct products (google the notes titled SEMIDIRECT PRODUCTS by KEITH CONRAD) should be known before you read the proof.
Z Wu
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