# Do all exact 1 -> A -> AxB -> B -> 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 $g((n_i),(m_i)) = (\overline{n_1}, m_1, \overline{n_2}, m_2, \ldots)$.

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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
@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

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