$$A\to B\to C\to D\to E$$ be an exact sequence with all groups (finitely generated) $2$-torsion abelian groups, except possibly $C$. Trivial groups are also permitted.

Does it follow that $C$ is also $2$-torsion?

My idea would be to tensor with $\Bbb F_p$ so then everything is $0$ except the $C$ in the middle. But the resulting sequence won't be exact anymore in general, so it's useless.


1 Answer 1


By taking the quotient of $B$ by the kernel of $B\to C$, we can rewrite this as a left exact sequence with $B'$ and $D$ $2$-torsion groups:

$$0\to B'\xrightarrow{f} C \xrightarrow{g} D$$

If $c\in C$, then, because $D$ is $2$-torsion, there is some $k$ with $2^k g(c) = 0$, so $2^k c\in\ker g$. Then $2^kc \in \operatorname{im} f$, so, because $f$ is injective and $B'$ is $2$-torsion, there is some $l$ with $2^l (2^k c) = 0$, so $2^{l+k} c=0$ and $c$ is $2$-torsion.


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