TITLE: If $M \simeq N$ in ${\tt stmod}(G)$ will $M \oplus \text{(proj)} \simeq N \oplus \text{(proj)}$ in ${\tt mod}(G)$?
QUESTION [5 upvotes]: Let $G$ be a finite group and ${\tt stmod}(G)$ the stable module category for $G$, i.e., the category whose objects are $G$-modules and whose morphisms are $G$-module homomorphisms modulo those that factor through a projective module.
Is it true that two modules $M$ and $N$ are isomorphic in the stable module category if and only if there are projectives $P$ and $Q$ and an isomorphism $M \oplus P \simeq N \oplus Q$ of $G$-modules?
It's easy to show that if $M \oplus P \simeq N \oplus Q$ then $M$ and $N$ are isomorphic in ${\tt stmod}(G)$.  But the other direction eludes me.  If $M$ and $N$ are isomorphic in ${\tt stmod}(G)$ then there are maps $\phi\colon M \to N$ and $\psi\colon N \to M$ such that $\phi\psi - \operatorname{id}_N$ and $\psi\phi - \operatorname{id}_M$ factor through projectives $P$ and $Q$ respectively.  I tried defining a homomorphisms between $M \oplus P$ and $N \oplus Q$ using these factorization maps but I can't figure out how to do it so that they're inverse to each other.

REPLY [6 votes]: Yes. In fact, there is a general theorem due to Heller, see The loop space functor in homological algebra Trans. Amer. Math. Soc. 96 (1960), 382-394, Theorem 2.2:

Suppose $\mathcal{A}$ is an additive category in which idempotents split. Let $\mathcal{P}$ be a class of objects in $\mathcal{A}$ such that $P \oplus Q$ is in $\mathcal{P}$ if and only if $P$ and $Q$ are in $\mathcal{P}$. Let $\langle \mathcal P \rangle$ be the ideal of morphisms of $\mathcal{A}$ factoring through an object of $\mathcal{P}$. 
Then two objects $A$ and $B$ of $\mathcal{A}$ become isomorphic in the quotient category
   $\mathcal{A}/\langle \mathcal{P}\rangle$ if and only if there are $P$
   and $Q$ in $\mathcal{P}$ such that $A \oplus P$ is isomorphic to $B
 \oplus Q$ in $\mathcal{A}$.

The proof is a slightly technical, but rather direct verification. Heller's argument is very clear, so I won't reproduce it.
See also this MO thread.