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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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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. –  Martin Brandenburg Nov 4 '11 at 9:13
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@Martin Brandenburg: at mathoverflow.net/questions/23478/… you write that "Every short exact sequence of [the form you mention] splits" is a false belief. –  aorq Nov 5 '11 at 4:09
    
@A. Rex: Hehe, that's true. –  Martin Brandenburg Nov 29 '11 at 8:42

1 Answer 1

up vote 32 down vote accepted

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

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What I find surprising is that this was only proved 5 years ago. –  Sándor Kovács Nov 4 '11 at 4:04
    
Just to clarify, the answer to the question in the title is true, not the question in the second paragraph. –  Peter Samuelson Nov 4 '11 at 4:23
    
Thank you Hailong for bringing this to my attention! Splitting of the sequence above on the right yields $A \oplus B$ as a (potentially non-trivial) semi-direct product. But in fact we get even more, namely that the sequence splits on the left, meaning that $A \oplus B$ is actually a direct product. –  Dan Glasscock Nov 4 '11 at 4:51

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