# Proving that some property on a chain complex of groups implies isomorphism between direct sums of these groups.

Let $C_*$ be a chain complex such that every $C_i$ is a torsion-free finitely generated abelian group, with $C_i=0$ for every $i<0$ and every $i>N$ for some sufficiently large integer $N$. If every homology group $H_i(C_*)$ is a torsion group, then $$\bigoplus_{i \text{ even}}C_i\cong\bigoplus_{i \text{ odd}}C_i.$$

I honestly have no clue of how to begin proving this. Any help would be appreciated

Since the $C_i$ are free, only the ranks matter, and we may tensor with $\mathbb{Q}$. Then the complex becomes exact because the homology was assumed to be torsion. Now use that the Euler characteristic of an exact complex vanishes.
• The key (the only!?) ingredient from this that you need at the level of f.g. abelian groups is that if $0\to A'\to A\to A''\to0$ is a short exact secuence, then the ranks of the free parts add up, i.e. $r(A)=r(A')+r(A'')$. Do this for all the SES involving chains, cycles, boundaries, homology. Massive cancellation will take place! Commented Sep 27, 2014 at 20:50