Let $H\!\leq\!G$ be finite groups and $C_\ast(H)\!\leq\!C_\ast(G)$ their bar complexes (each $C_k(G)$ is a free $R$-module with basis $(G\!\setminus\!\{1\})^k$).
Is it possible that $H$ is not a direct summand, but $C_\ast(H)$ is, i.e. that there holds $$G\ncong H\oplus\ldots~~\text{ and }~~C_\ast(G)\cong C_\ast(H)\oplus\ldots?$$
Furthermore, how about $~G\ncong H\oplus\ldots~~$ and $~~C_\ast(G)\simeq C_\ast(H)\oplus\ldots?$
Cyclic and dihedral groups have $D_n\ncong C_n\oplus C_2$, but homology shows $H_\ast D_n\ncong H_\ast C_n\oplus\ldots$.