Representation theory

Linear algebra alone guarantees some complement to $W$, but a generic complement will not be $G$-invariant. Maschke's theorem shows we can always average an arbitrary projection over the group to force it to commute with the $G$-action, at the cost of dividing by $|G|$ — which is exactly why the theorem needs $\operatorname{char}(k) \nmid |G|$, and exactly why modular representation theory (where this division is impossible) behaves so differently.

Step 1 (start with any complement). By ordinary linear algebra, choose any vector-space complement $U$ to $W$ in $V$ (not necessarily $G$-invariant), and let $\pi: V \to W$ be the corresponding projection ($\pi|_W = \mathrm{id}$, $\ker\pi = U$).

Step 2 (average over the group). Define $\bar\pi = \dfrac{1}{|G|}\sum_{g \in G} \rho(g)\, \pi\, \rho(g)^{-1}$. This makes sense exactly because $\operatorname{char}(k) \nmid |G|$, so $|G|$ is invertible in $k$.

Step 3 ($\bar\pi$ is still a projection onto $W$). For $w \in W$, since $W$ is $G$-invariant, $\rho(g)^{-1}w \in W$, so $\pi(\rho(g)^{-1}w) = \rho(g)^{-1}w$, giving $\rho(g)\pi(\rho(g)^{-1}w) = w$ for every term; averaging these $|G|$ copies of $w$ gives $\bar\pi(w) = w$. For general $v \in V$, $\pi(\rho(g)^{-1}v) \in W$ always (image of $\pi$), so $\rho(g)\pi(\rho(g)^{-1}v) \in \rho(g)W = W$ by invariance, and averaging stays in $W$; hence $\bar\pi: V \to W$ is a projection onto $W$.

Step 4 ($\bar\pi$ commutes with the action, so its kernel is $G$-invariant). For any $h \in G$, $\rho(h)\bar\pi\rho(h)^{-1} = \dfrac{1}{|G|}\sum_{g} \rho(hg)\pi\rho(hg)^{-1} = \bar\pi$ since $g \mapsto hg$ merely permutes the sum; so $\rho(h)\bar\pi = \bar\pi\rho(h)$. Setting $W' = \ker\bar\pi$, for $v \in W'$ we get $\bar\pi(\rho(h)v) = \rho(h)\bar\pi(v) = \rho(h)(0) = 0$, so $\rho(h)v \in W'$: $W'$ is $G$-invariant, and $V = W \oplus W'$ since $\bar\pi$ is a projection onto $W$. Iterating this splitting on $W'$ (finite dimension forces termination) decomposes $V$ into a direct sum of irreducibles.

Schur's Lemma says irreducible representations are as rigid as possible: the only maps between them that respect the group action are either zero or invertible, and self-maps are just scalars. This rigidity is exactly what forces the orthogonality relation, which in turn gives an extremely practical tool — a dot-product test on a finite table of numbers (the character table) that instantly tells you which representations are irreducible and how any representation decomposes.

Step 1 (Schur's Lemma, kernel and image). Let $f: V \to W$ be $G$-equivariant, i.e. $f(\rho_V(g)v) = \rho_W(g)f(v)$ for all $g,v$. Then $\ker f \subseteq V$ is a $G$-invariant subspace (if $f(v)=0$ then $f(\rho_V(g)v) = \rho_W(g)f(v) = 0$), and since $V$ is irreducible, $\ker f$ is either $0$ or all of $V$.

Step 2 (conclude $f=0$ or $f$ injective, then surjective). If $\ker f = V$, then $f = 0$. Otherwise $\ker f = 0$, so $f$ is injective; the image $f(V) \subseteq W$ is also $G$-invariant (by equivariance), and since $W$ is irreducible and $f(V) \ne 0$ (as $f\ne0$), $f(V) = W$, making $f$ bijective, hence an isomorphism.

Step 3 ($V=W$ forces $f=\lambda I$). Since $\mathbb{C}$ is algebraically closed, $f$ has an eigenvalue $\lambda$ with eigenspace $E_\lambda = \ker(f-\lambda I) \ne 0$. Because $f$ is $G$-equivariant, so is $f - \lambda I$, so $E_\lambda$ is a nonzero $G$-invariant subspace of the irreducible $V$, forcing $E_\lambda = V$; hence $f = \lambda I$.

Step 4 (orthogonality from averaging). Fix bases and let $f: V_j \to V_i$ be an arbitrary linear map; form the equivariant map $\bar f = \dfrac{1}{|G|}\sum_{g} \rho_i(g) f \rho_j(g)^{-1}$ exactly as in Maschke's theorem. By Steps 1–3, $\bar f = 0$ if $V_i \not\cong V_j$, and $\bar f = \lambda I$ (with $\lambda$ computable via trace) if $V_i = V_j$. Choosing $f$ to be elementary matrix units and comparing matrix entries of $\bar f$ on both sides of this identity yields, after summing over the diagonal to extract traces, exactly $\dfrac{1}{|G|}\sum_{g \in G} \chi_i(g)\overline{\chi_j(g)} = \delta_{ij}$.