In Carlson's Cohomology and representation theory, the author states Schanuel's lemma, and then derives a consequence that I cannot understand.
They define, for a $kG$ module $M$, $\Omega (M)$ to be the kernel of a surjective map $P\to M$ where $P$ is projective of minimal dimension (here I'm assuming dimension means "dimension over $k$" ); and then state that if $\gamma : Q\to M$ is another surjective map from a projective, then $\ker \gamma = \Omega( M) \oplus (\mathrm{proj})$, and say that it follows from Schanuel's lemma.
But I don't see how that follows from Schanuel's lemma. The exact statement is that $\ker \gamma \oplus P \cong \Omega(M)\oplus Q$, but I don't see why the isomorphism should take $\ker\gamma$ to $\Omega(M)\oplus$some projective submodule of $Q$ (actually the isomorphism I found when I wrote down a proof of the lemma doesn't send $\Omega(M)$ into $\ker \gamma$ at all)
The ingredient you're missing is the Krull-Schmidt theorem, which says that if you decompose a finite length module over a ring as a direct sum of indecomposable modules, the summands are unique up to isomorphism (and reordering). In particular, let's decompose each term of $$\ker \gamma \oplus P \cong \Omega(M)\oplus Q$$ into indecomposable summands. Since $\Omega(M)$ has no projective summands and $Q$ has only projective summands, we see that all the non-projective indecomposable summands of $\ker \gamma$ and $P$ must correspond to summands of $\Omega(M)$ and all the projective indecomposable summands of $\ker \gamma$ and $P$ must correspond to summands of $Q$. But since $P$ is projective, all the non-projective summands must come from $\ker\gamma$. This means that the decomposition of $\ker\gamma$ consists of all the summands of $\Omega(M)$ plus possibly some of the summands of $Q$. That is, $\ker\gamma\cong \Omega(M)\oplus Q'$ for some projective module $Q'$.