# Do all exact $1 \to A \to A \times B \to B \to 1$ split for finite groups?

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 $g((n_i),(m_i)) = (\overline{n_1}, m_1, \overline{n_2}, m_2, \ldots)$.

• 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
• @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

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.

• 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
• 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. – YCor Feb 12 '16 at 17:31
• 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. – zibadawa timmy Oct 17 '17 at 11:33