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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?

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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 $\mathbb{Z} \oplus \mathbb{Z}_2^{\mathbb{N}}\to\mathbb{Z}$.

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  • $\begingroup$ Do they both need to be the canonical injection/projection, or just one of them? In my situation, sequence comes from $H\cong G/F$, so the maps are the canonical injection and projection. I this correct? $\endgroup$
    – Felix Y.
    May 11, 2014 at 18:02
  • $\begingroup$ If the maps are the canonical maps given by the quotient construction then the sequence splits by the splitting lemma (and in fact the converse holds as well). $\endgroup$
    – Dan Rust
    May 11, 2014 at 18:08
  • $\begingroup$ About your example in your edit, what if all of your groups are finitely generated abelian groups? $\endgroup$
    – Felix Y.
    May 11, 2014 at 18:22
  • $\begingroup$ I think finitely generated may be enough to ensure a splitting. I'll try and find a reference. $\endgroup$
    – Dan Rust
    May 11, 2014 at 18:23
  • $\begingroup$ There appears to be an answer here and in fact the question uses the same example as I do for an example where the sequence doesn't split if you allow infinitely generated groups. $\endgroup$
    – Dan Rust
    May 11, 2014 at 18:24

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