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Let $A$, $B$ be finite groups. Is it true that all short exact sequences $1 \rightarrow A \rightarrow A \times B \rightarrow B \rightarrow 1$ split on the right?

In other words, do there exist finite groups $A$, $B$ and homomorphisms $f: A \rightarrow A \times B$, $g: A \times B \rightarrow B$ such that $1 \rightarrow A \rightarrow A \times B \rightarrow B \rightarrow 1$ is exact and there does not exist a homomorphism $h: B \rightarrow A \times B$ such that $g \circ h = \text{id}_B$?

An example when $A$, $B$ are not finite is given by $A = \prod_{i=1}^\infty \mathbb{Z}$, $B = \prod_{i=1}^\infty \mathbb{Z}/2\mathbb{Z}$, $f((n_i)) = ((2n_i),0)$, and $g((n_i),(m_i)) = (\overline{n_1}, m_1, \overline{n_2}, m_2, \ldots)$.

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    $\begingroup$ Minor comment: Writing down the sequence $1 \to A \to A \times B \to B \to 1$ without specification of the morphisms always means that we consider the inclusion $A \to A \times B$ and the projection $A \times B \to B$. But anyway, you have clarified this in the second paragraph. $\endgroup$ Commented Nov 4, 2011 at 9:13
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    $\begingroup$ @Martin Brandenburg: at mathoverflow.net/questions/23478/… you write that "Every short exact sequence of [the form you mention] splits" is a false belief. $\endgroup$
    – aorq
    Commented Nov 5, 2011 at 4:09
  • $\begingroup$ @A. Rex: Hehe, that's true. $\endgroup$ Commented Nov 29, 2011 at 8:42

1 Answer 1

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This is true (1). It was extended to finitely generated profinite groups here (2). Surprisingly, it is also true in the category of finitely generated modules over a Noetherian commutative ring (3).

(1) Joseph Ayoub, The direct extension theorem, J. Group Theory 9 (2006) 307-316.

(2) Goldstein, Daniel, Guralnick, Robert, The direct product theorem for profinite groups. J. Group Theory 9 (2006), no. 3, 317-322.

(3) Takehiko Miyata, Note on direct summands of modules. J. Math. Kyoto Univ. Volume 7, Number 1 (1967), 65-69.

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    $\begingroup$ What I find surprising is that this was only proved 5 years ago. $\endgroup$ Commented Nov 4, 2011 at 4:04
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    $\begingroup$ Just to clarify, the answer to the question in the title is true, not the question in the second paragraph. $\endgroup$ Commented Nov 4, 2011 at 4:23
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    $\begingroup$ Could you write down the references? One of the links is not working; after inquiry it's: Goldstein, Daniel, Guralnick, Robert The direct product theorem for profinite groups. J. Group Theory 9 (2006), no. 3, 317-322. $\endgroup$
    – YCor
    Commented Feb 12, 2016 at 17:31
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    $\begingroup$ It may also be worth noting that Ayoub's proof also applies to any group satisfying both chain conditions with only minor changes. The only place his proof actually needs finiteness, rather than the analogous arguments with chain conditions, specifically is in results regarding the center $Z(G)$ and the abelianization $G/G'$, but these inherit the chain conditions and any abelian group satisfying both chain conditions is automatically finite. $\endgroup$ Commented Oct 17, 2017 at 11:33
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    $\begingroup$ Your hyperlink to (3) on Project Euclid is broken. $\endgroup$
    – KConrad
    Commented Apr 23, 2022 at 18:46

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