# Example of a non-splitting exact sequence $1 \to K \to G \to H \to 1$

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 $0 → M → M\oplus N → N → 0$).

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

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

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.]

• $\mathbb Z_2$ is not infinite. Oct 31, 2022 at 13:23
• 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. Oct 31, 2022 at 13:23
• I guess I didn't read the question very carefully. :P Oct 31, 2022 at 13:27
• Thank you, I don't think $Im(f) = ker(g)$ in this case? Oct 31, 2022 at 15:22
• The first group should be $\mathbb{Z}$ Oct 31, 2022 at 16:20