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Suppose $G$, $K$ and $H$ are finitely generated infinite groups such that
$$1 \to K \to G \to H \to 1$$ is a short exact sequence.

Question: Are there examples where this sequence does not split?

There are examples of such groups when $H$ is finite(Does every short exact sequence split?) or when $G$ is not a finitely generated group (Example of a non-splitting exact sequence $0 → M → M\oplus N → N → 0$).

Thank you for your help.

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    $\begingroup$ $1 \to \mathbb Z \times C_2 \to \mathbb Z^2 \times C_4 \to \mathbb Z \times C_2 \to 1$. $\endgroup$ Oct 31, 2022 at 13:17

1 Answer 1

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$0 \to \mathbb{Z}^2 \to \mathbb{Z}^2 \to \mathbb{Z}\oplus\mathbb{Z}_2 \to 0$

using maps $(x,y) \mapsto (x,2y)$ and $(x,y) \to (x,y\text{ mod }2)$.

The only morphism from $\mathbb{Z}_2$ to $\mathbb{Z}$ is trivial, so you cannot split.

In more detail: Say $f(x,y) = (x,y\text{ mod }2)$ is our map from $\mathbb{Z}^2 \to \mathbb{Z}\oplus \mathbb{Z}_2$. Now $\mathbb{Z}^2$ doesn't have any elements of order 2, so to have a morphism $g:\mathbb{Z} \oplus \mathbb{Z}_2 \to \mathbb{Z}^2$, you are forced to send $g(0,1)$ to $(0,0)$. Thus $g \circ f$ cannot be the identity. [There's no split.]

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    $\begingroup$ $\mathbb Z_2$ is not infinite. $\endgroup$ Oct 31, 2022 at 13:23
  • $\begingroup$ To be fair, they ask for all three groups to be infinite. To fix that, you can just add an extra factor of $\mathbb{Z}$ to each group, and define the maps to be the identity on those components. $\endgroup$ Oct 31, 2022 at 13:23
  • $\begingroup$ I guess I didn't read the question very carefully. :P $\endgroup$
    – Bill Cook
    Oct 31, 2022 at 13:27
  • $\begingroup$ Thank you, I don't think $Im(f) = ker(g)$ in this case? $\endgroup$
    – ghc1997
    Oct 31, 2022 at 15:22
  • $\begingroup$ The first group should be $\mathbb{Z}$ $\endgroup$ Oct 31, 2022 at 16:20

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