What is the defining condition for a group-invariant inner product in the proof of Maschke's theorem?

The block-diagonal matrix form represents the decomposition of the representation $\rho$ into independent subrepresentations on the invariant subspaces $W$ and $W^{\perp}$. When $V = W \oplus W^{\perp}$ and both summands are $G$-invariant, the matrix of $\rho(g)$ in a basis adapted to this direct sum has zero off-diagonal blocks, meaning the action on $W$ does not mix with the action on $W^{\perp}$.

Conditions: $V = W \oplus W^{\perp}$ is a direct sum decomposition.; $W$ and $W^{\perp}$ are both $G$-invariant subspaces.; A basis for $V$ is chosen as the union of bases for $W$ and $W^{\perp}$.

If $W$ is a $G$-invariant subspace and the inner product is $G$-invariant, then for any $y \in W^{\perp}$ and $g \in G$, the vector $\rho(g)y$ remains orthogonal to every $w \in W$. This is shown by rewriting $(\rho(g)y, w)$ as $(y, \rho(g^{-1})w)$.

Conditions: $W \subseteq V$ is a $G$-invariant subspace ($\rho(g)W \subseteq W$ for all $g \in G$).; The inner product $(\cdot, \cdot)$ on $V$ is $G$-invariant ($(\rho(g)v, \rho(g)w) = (v, w)$).; $W^{\perp} = \{y \in V \mid (y, w) = 0 \text{ for all } w \in W\}$.

The decomposition terminates because the dimension of the representation space is finite. At each step, a proper nonzero $G$-invariant subspace $W$ is split off, leaving a complementary invariant subspace $W^{\perp}$.

Conditions: $V$ is a finite-dimensional vector space.; $W$ is a proper nonzero $G$-invariant subspace of $V$.; The inner product is $G$-invariant, ensuring $W^{\perp}$ is also $G$-invariant.