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I am learning group extension following Weibel's An Introduction to Homological Algebra, and there is an exercise in the book: $A$ is an abelian group, and $G$ is a group.

Show that an extension $0\to A\to E\xrightarrow[]\pi G\to 1$ splits if and only if $E$ is isomorphic to the semidirect product $A\rtimes G$.

I have proved why $E\cong A\rtimes G$ if the extension splits, but I can't figure out how to prove the other direction.

The main trouble is: What's the relation of the semidirect product and the original map?

Given an isomorphism $\phi:A\rtimes G\to E$, is it true that $\pi\circ \phi=\pi'$? ($\pi: E\to G$ is the original map, and $\pi': A\rtimes G\to G, (a,g)\mapsto g$ the natural map).

To be clear, given an extension $0\to A\to E\xrightarrow[]\pi G\to 1$, the definition of $A$ as a $G$-mod is: for $g\in G$ and $a\in A$, take $\tilde g\in \pi^{-1}(g),$ define $ g \cdot a=\tilde ga\tilde g^{-1}$, which is well defined as the abelian group $A$ is identified with $\ker (E\xrightarrow[]\pi G)$.

Moreover, an extension $0\to A\to E\xrightarrow[]\pi G\to 1$ splits if there exists section $\sigma:G\to E$ such that $\pi\circ \sigma=\mathrm{id}_G$.

shwsq
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  • This may help: https://math.stackexchange.com/questions/339731/what-is-the-relation-between-semidirect-products-extensions-and-split-extensio?rq=1 – FShrike May 06 '25 at 11:13
  • @FShrike But in that answer, it proves only that there exists a short exact sequence of groups $0\to A\to E\to G\to 1$ that splits, while I need to prove that a given exact sequence splits. – shwsq May 06 '25 at 12:17
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    Maybe the statement you are trying to prove is wrong (I haven't got a copy of Weibel's book so I don't know if it's wrong in the book). The sequence splits if and only if there is an automorphism of $E$ with $A\rtimes G$ such that this diagram commutes.

    $$ \begin{array}{ccc} & E & \ \stackrel{\scriptstyle\pi}{\swarrow} & & \searrow \ G & \leftarrow & A \rtimes G \end{array} $$

    If you only know that there is some isomorphism of $E$ with the semidirect product, but not that this isomorphism makes the above diagram commute, it's possible that the given sequence doesn't split.

     – hunter May 06 '25 at 13:31
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    I'm not able to come up with a counterexample, though. – hunter May 06 '25 at 13:31
  • Thanks a lot. Indeed there are counterexamples if we only have the isomorphism $E\cong A\rtimes G$, for example in https://math.stackexchange.com/questions/135444/a-nonsplit-short-exact-sequence-of-abelian-groups-with-b-cong-a-oplus-c. – shwsq May 06 '25 at 14:14

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The suggestive lingo "splits" implies that the extension $E$ can be written as a type (semi-direct) of product.

However the converse turns out to be false. See the link in the comments. Incidentally it looks like it takes a little work to get a counterexample.

Thus semi-direct product generalizes the short exact sequence splitting.

suckling pig
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