If I have an exact sequence of abelian groups, the sequence coming from knowing that $H\cong G/F$,
$$0\rightarrow F \rightarrow G \rightarrow H \rightarrow 0$$
where I know that $ F\oplus H\subset G$, does that mean it splits?
What if all of the groups are also finitely generated?
Best Answer
Not necessarily, you also need that the maps are (from left to right) the canonical injection and the canonical projection respectively (up to isomorphism).
As an example where the sequence doesn't split, consider
$$0 \to \mathbb{Z} \stackrel{f}{\to} \mathbb{Z} \oplus \mathbb{Z}_2^{\mathbb{N}} \stackrel{g}{\to} \mathbb{Z}_2^{\mathbb{N}} \to 0$$ and set $f(n)=(2n,0,0,0,\ldots)$ and $g(n,m_1,m_2,m_3\ldots)=([n],m_1,m_2,m_3,\dotsc)$.
Notice that $g$ is not the canonical projection as the kernel of $g$ is not the whole subgroup $\langle(1,0,0,0,\ldots)\rangle$ but only $\langle(2,0,0,0,\ldots)\rangle$. Similarly with $f$. The sequence does not split as the only homomorphisms from $\mathbb{Z}_2^{\mathbb{N}}$ to $\mathbb{Z} \oplus \mathbb{Z}_2^{\mathbb{N}}$ have zero image when composed with the projection map $\mathbb{Z} \oplus \mathbb{Z}_2^{\mathbb{N}}\to\mathbb{Z}$.