Maschke's theorem

Theorem

Maschke's theorem

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Statement

Let \(G\) be a finite group and let \(\rho: G \to GL(V)\) be a representation of \(G\) on a finite-dimensional vector space \(V\) over \(\mathbb{C}\) (a field of characteristic \(0\) suffices; more generally, any field \(k\) whose characteristic does not divide \(|G|\)). If \(W \subseteq V\) is a \(G\)-invariant subspace, then there exists a \(G\)-invariant subspace \(W' \subseteq V\) such that \(V = W \oplus W'\) as \(G\)-representations. Equivalently: every finite-dimensional representation of \(G\) over \(\mathbb{C}\) is completely reducible, i.e. decomposes as a direct sum of irreducible subrepresentations.

Why it matters

Maschke's theorem is the load-bearing structural fact underlying the entire representation theory of finite groups. It converts the study of an arbitrary representation into the study of a finite list of building blocks — the irreducibles — because complete reducibility means every representation is, up to isomorphism, a direct sum of irreducibles with multiplicities. Without it, the category of \(G\)-representations would have indecomposable objects that are not irreducible (as happens for modular representations, char \(k \mid |G|\)), and character theory as a diagonalisation tool would collapse.

The theorem is also the precise reason the group algebra \(\mathbb{C}G\) is semisimple: Maschke's theorem for the regular representation, combined with the Artin–Wedderburn structure theorem, is what delivers \(\mathbb{C}G \cong \bigoplus_i \mathrm{Mat}_{n_i}(\mathbb{C})\), the engine behind character orthogonality relations.

Hypotheses

\(G\) is finite. Counterexample if dropped: let \(G = \mathbb{Z}\) act on \(V = \mathbb{C}^2\) by \(n \cdot (x,y) = (x, y+nx)\) (unipotent shear representation). The line \(W = \{(x,0)\}\) is invariant, but any complementary line is moved by large \(n\); no invariant complement exists, so \(V\) is indecomposable but not irreducible.

The ground field has characteristic \(0\), or at least \(\operatorname{char}(k) \nmid |G|\). Counterexample if dropped: let \(G = \mathbb{Z}/2\mathbb{Z} = \{1,\sigma\}\) act on \(V = k^2\) over \(k = \mathbb{F}_2\) by \(\sigma \cdot (x,y) = (x, x+y)\) (a nontrivial Jordan block, since \(2 = |G|\) vanishes in \(k\)). The line \(W=\{(0,y)\}\) is invariant; every other line is \(\{(x,cx)\}\), and \(\sigma\) sends this to \(\{(x, x+cx)\} = \{(x,(1+c)x)\}\), equal to itself only if \(1+c=c\), impossible in \(\mathbb{F}_2\). No invariant complement to \(W\) exists.

\(V\) is finite-dimensional (or more generally the averaging/integral below converges and the invariant form is well defined). The finite-dimensionality is used only to guarantee \(W\) has a subspace complement at all (linear algebra) and that traces/averages are finite sums; the theorem extends verbatim to infinite-dimensional \(V\) using the same averaged-projection argument, and separately to compact groups \(G\) replacing the group average \(\frac{1}{|G|}\sum_{g}\) by the Haar integral \(\int_G dg\) — the finiteness of \(G\) is really only needed to make "\(\frac{1}{|G|}\)" meaningful.

Proof

\text{Since } V \text{ is finite-dimensional, choose (by ordinary linear algebra) any vector-space complement } U \text{ of } W: \quad V = W \oplus U.

Every subspace of a finite-dimensional vector space has a complement; this uses no group structure yet — only that \(V/W\) is finite-dimensional and we can lift a basis. A

\text{Let } \pi: V \to W \text{ be the projection with } \ker \pi = U \text{ and } \pi|_W = \mathrm{id}_W.

Direct sum decompositions correspond bijectively to idempotent linear projections onto one summand; \(\pi\) is \(k\)-linear by construction but need not commute with the \(G\)-action yet. A

\text{Define } \tilde\pi := \frac{1}{|G|}\sum_{g \in G} \rho(g)\, \pi\, \rho(g)^{-1} : V \to V.

This is the averaging (Reynolds operator) construction: symmetrising a non-invariant map over the finite group \(G\) to force invariance. Division by \(|G|\) is legal because \(\operatorname{char}(k) \nmid |G|\) (or \(k = \mathbb{C}\)), so \(|G|\) is invertible in \(k\). B

\tilde\pi(V) \subseteq W.

For each \(g\), \(\pi\rho(g)^{-1}v \in W\) since \(\pi\) has image \(W\), and \(\rho(g)\) preserves \(W\) because \(W\) is a \(G\)-invariant subspace by hypothesis; so each summand \(\rho(g)\pi\rho(g)^{-1}v\) lies in \(W\), and \(W\) is closed under the sum and the scalar \(1/|G|\). A

\text{For } w \in W: \quad \tilde\pi(w) = \frac{1}{|G|}\sum_{g\in G} \rho(g)\,\pi\big(\rho(g)^{-1}w\big) = \frac{1}{|G|}\sum_{g\in G} \rho(g)\rho(g)^{-1}w = w.

Since \(W\) is \(G\)-invariant, \(\rho(g)^{-1}w \in W\) for every \(g\), so \(\pi\) fixes it (as \(\pi|_W = \mathrm{id}_W\) from Step 2); the \(|G|\) copies of \(w\) average back to \(w\). Hence \(\tilde\pi|_W = \mathrm{id}_W\). A

\tilde\pi \text{ is a linear projection onto } W \quad (\tilde\pi^2 = \tilde\pi), \text{ since } \tilde\pi(V) \subseteq W \text{ and } \tilde\pi|_W = \mathrm{id}_W.

A linear map that maps into a subspace and restricts to the identity there is automatically idempotent: for \(v \in V\), \(\tilde\pi(v) \in W\) so \(\tilde\pi(\tilde\pi(v)) = \tilde\pi(v)\) by Step 5. A

\tilde\pi \text{ is } G\text{-equivariant}: \quad \rho(h)\,\tilde\pi\,\rho(h)^{-1} = \tilde\pi \quad \text{for all } h \in G.

Compute directly: \(\rho(h)\tilde\pi\rho(h)^{-1} = \frac{1}{|G|}\sum_{g\in G} \rho(hg)\,\pi\,\rho(hg)^{-1} = \frac{1}{|G|}\sum_{g'\in G} \rho(g')\,\pi\,\rho(g')^{-1} = \tilde\pi\), reindexing \(g' = hg\), which is a bijection of \(G\) with itself since \(G\) is a group (left multiplication by \(h\) is a bijection \(G \to G\)). This is precisely the averaging trick's payoff. C

\text{Set } W' := \ker \tilde\pi. \text{ Then } W' \text{ is a } G\text{-invariant subspace and } V = W \oplus W'.

Kernel of a linear map is a subspace; \(V = \mathrm{im}(\tilde\pi) \oplus \ker(\tilde\pi) = W \oplus W'\) is the standard decomposition induced by any idempotent (Step 6), together with \(\mathrm{im}(\tilde\pi)=W\) from Steps 4–5. Invariance of \(W'\) under \(G\) follows from Step 7: if \(\tilde\pi(v) = 0\) then \(\tilde\pi(\rho(h)v) = \rho(h)\tilde\pi\rho(h)^{-1}\rho(h)v = \rho(h)\tilde\pi(v) = 0\), so \(\rho(h)v \in W' = \ker\tilde\pi\). B

\text{By induction on } \dim V: \quad \text{every finite-dimensional representation of } G \text{ over } \mathbb{C} \text{ is a direct sum of irreducibles.}

If \(V\) is irreducible, done. Otherwise \(V\) has a proper nonzero invariant subspace \(W\); by Steps 1–8 \(V = W \oplus W'\) with both summands of strictly smaller dimension, so by the inductive hypothesis (base case \(\dim V = 0\) or \(1\), trivially irreducible/zero) each of \(W, W'\) decomposes into irreducibles, hence so does \(V\). B

Result

V = W \oplus W' \quad \text{for some } G\text{-invariant } W', \text{ whenever } W \leq V \text{ is } G\text{-invariant and } \operatorname{char}(k) \nmid |G|

Reading. You never get "stuck" with an invariant subspace that has no invariant partner: the averaging trick \(\tilde\pi = \frac{1}{|G|}\sum_g \rho(g)\pi\rho(g)^{-1}\) manufactures a genuinely equivariant projection out of any ordinary linear one, and its kernel is the complement you want. Iterating, every representation splits completely into irreducible pieces — there are no "extension" phenomena to worry about, unlike in modular representation theory.

Scope. Holds for any finite group \(G\) and any field \(k\) with \(\operatorname{char}(k) \nmid |G|\) (in particular \(k = \mathbb{C}, \mathbb{R}, \mathbb{Q}\), or any finite field \(\mathbb{F}_q\) with \(\gcd(q,|G|)=1\)); fails in general when \(\operatorname{char}(k) \mid |G|\) (modular representation theory) and fails for infinite discrete \(G\) unless replaced by suitable compactness (e.g. compact Lie groups via Haar measure).

Corollaries & converses

Every finite-dimensional \(\mathbb{C}G\)-module is a direct sum of simple modules, i.e. \(\mathbb{C}G\) is a semisimple algebra (Maschke + Artin–Wedderburn \(\Rightarrow\) \(\mathbb{C}G \cong \bigoplus_i \mathrm{Mat}_{n_i}(\mathbb{C})\)).
Any \(G\)-invariant Hermitian inner product exists on \(V\) (average any inner product over \(G\)), and complete reducibility can equivalently be proved by taking \(W'\) to be the orthogonal complement of \(W\) with respect to such an averaged inner product — this is the "unitary trick" version of the same argument.
Two representations with the same character are isomorphic (relies on complete reducibility to reduce isomorphism-checking to multiplicity-checking against irreducibles).
Converse: the converse "complete reducibility of all representations of \(G\) over \(k\) implies \(\operatorname{char}(k) \nmid |G|\)" is true and is essentially Maschke's theorem run backwards: if \(p = \operatorname{char}(k)\) divides \(|G|\), the regular representation \(kG\) itself is not completely reducible (the augmentation ideal fails to split off a trivial complement), so complete reducibility for all \(G\)-modules over \(k\) exactly characterises \(p \nmid |G|\).

Fails without

Infinite group, no compactness: \(G = \mathbb{Z}\) acting on \(\mathbb{C}^2\) by \(n \mapsto \begin{pmatrix}1 & n\\ 0 & 1\end{pmatrix}\); the invariant line \(W = \mathbb{C} e_1\) has no invariant complement (any other line is dragged off by large \(n\)), so this representation is indecomposable but reducible — Maschke's conclusion fails outright.
Characteristic dividing the group order: \(G = \mathbb{Z}/p\mathbb{Z}\) acting on \(V = \mathbb{F}_p^2\) by the regular representation, realised via the Jordan block \(\begin{pmatrix}1&1\\0&1\end{pmatrix}\) for a generator; the line \(W\) spanned by \((1,0)\) is invariant but every candidate complement line is moved by the generator (the matrix is a nontrivial unipotent, not diagonalisable over \(\mathbb{F}_p\)), so no invariant complement exists.
Averaging breaks down explicitly: in the \(p \mid |G|\) case above, the "proof" formula \(\tilde\pi = \frac{1}{|G|}\sum_g \rho(g)\pi\rho(g)^{-1}\) is not merely unproven but ill-defined, since \(|G| \equiv 0\) in \(k\) and is not invertible — there is no division by \(|G|\) to perform, which is exactly where the argument structurally requires the hypothesis.

Common errors

Believing Maschke's theorem holds over any field — students forget the characteristic condition and misapply it to modular representations (e.g. representations of \(S_n\) over \(\mathbb{F}_p\) with \(p \leq n\)), then wrongly conclude modules there are semisimple.
Constructing \(\pi\) but forgetting to average it — using the ordinary (non-equivariant) projection \(\pi\) itself as "the" invariant complement's projection, which need not commute with the \(G\)-action at all.
Sloppy reindexing in the equivariance computation (Step 7) — asserting \(\sum_g \rho(hg)(\cdots)\rho(hg)^{-1} = \sum_g \rho(g)(\cdots)\rho(g)^{-1}\) without justifying that \(g \mapsto hg\) is a bijection of \(G\).
Confusing "completely reducible" with "irreducible" — Maschke gives a direct sum of irreducibles, not that \(V\) itself is irreducible; students sometimes state the conclusion as "\(V\) is irreducible."
Thinking the invariant complement \(W'\) is unique — it generally is not (any \(G\)-invariant subspace transverse to \(W\) works), though the isomorphism class of the resulting decomposition into irreducible summands, with multiplicities, is unique by Jordan–Hölder / uniqueness of semisimple decomposition.

Discussion

Heinrich Maschke proved this result in 1899, originally in the language of matrix groups rather than modules, at a time when representation theory was just being founded by Frobenius. The theorem's real content is algebraic: it says the group algebra \(kG\) is a semisimple ring exactly when \(\operatorname{char}(k) \nmid |G|\), and semisimple rings are completely classified by the Artin–Wedderburn theorem as products of matrix algebras over division rings. For \(k = \mathbb{C}\), those division rings are automatically \(\mathbb{C}\) itself (by a theorem of Frobenius/Schur on algebraically closed fields), giving \(\mathbb{C}G \cong \bigoplus_i \mathrm{Mat}_{n_i}(\mathbb{C})\) with \(\sum_i n_i^2 = |G|\) — the numerical identity behind orthogonality of characters.

The averaging argument used here — replace a non-canonical linear construction by its group-average to force equivariance — is one of the most versatile tricks in mathematics, reappearing as the "unitary trick" (Weyl) for compact Lie groups replacing \(\frac{1}{|G|}\sum_g\) with \(\int_G dg\) against Haar measure, and in invariant theory (Reynolds operators) for reductive algebraic groups. Complete reducibility for compact Lie groups (e.g. \(SO(3)\), \(SU(2)\)) is proved by literally the same computation with a sum replaced by an integral.

A subtler point: Maschke's theorem is a statement about the specific field/group pair, not merely "finite groups are nice." The modular representation theory of finite groups (Brauer, R. Brauer's theory) exists precisely because the conclusion fails when \(p \mid |G|\); the resulting category of \(kG\)-modules is not semisimple, has nontrivial Ext groups between simples, and projective and simple modules diverge — an entire parallel theory (blocks, defect groups, decomposition matrices) exists to organise the failure of Maschke's theorem.

Common misconception: that Maschke's theorem says every representation of a finite group is irreducible — it says the opposite in spirit: reducible representations are common, but they are always completely reducible (a direct sum of irreducibles), never merely "reducible with no complement," which is the pathology that can occur once either hypothesis is dropped.

Worked examples

G = \mathbb{Z}/3\mathbb{Z} = \{0,1,2\}, \quad \rho(1) = \begin{pmatrix}1&1\\0&1\end{pmatrix} \text{ acting on } V=\mathbb{C}^2 \text{... but first check this is genuinely a representation.}

Sanity check before applying the theorem: is \(\rho\) actually a homomorphism \(G \to GL_2(\mathbb{C})\)? We need \(\rho(1)^3 = I\). A

\begin{pmatrix}1&1\\0&1\end{pmatrix}^3 = \begin{pmatrix}1&3\\0&1\end{pmatrix} \neq I,

direct matrix multiplication of unipotent shears adds the off-diagonal entries; so this \(\rho\) is not a valid representation of \(\mathbb{Z}/3\mathbb{Z}\) over \(\mathbb{C}\) — it illustrates why the earlier "fails without" counterexample needed \(\mathbb{F}_p\) coefficients and \(G=\mathbb{Z}/p\mathbb{Z}\), where \(p \cdot 1 = 0\) rescues \(\rho(1)^p = I\). Over \(\mathbb{C}\) no such unipotent representation of a finite cyclic group exists — consistent with Maschke, since finite-order matrices over \(\mathbb{C}\) are always diagonalisable. Let us instead take a genuine representation. B

\text{Let } G=\mathbb{Z}/3\mathbb{Z}, \quad V = \mathbb{C}G \text{ (regular representation)}, \quad \rho(g)(e_h) = e_{gh}.

The regular representation of any finite group is always a valid representation, of dimension \(|G|\); we use it because it is the canonical place complete reducibility is applied to derive character orthogonality. A

W = \mathbb{C}\cdot(e_0+e_1+e_2) \quad \text{is } G\text{-invariant (the trivial subrepresentation)}.

\(\rho(g)(e_0+e_1+e_2) = e_g + e_{g+1} + e_{g+2} = e_0+e_1+e_2\) since \(g\mapsto g+k\) permutes \(\{0,1,2\}\); so \(W\) is fixed pointwise, hence invariant. A

\tilde\pi(v) = \frac{1}{3}\sum_{g} \rho(g)\pi\rho(g)^{-1}(v), \qquad \pi(e_h) := \tfrac{1}{3}\big[\text{coefficient matching}\big] \;\Rightarrow\; \tilde\pi(e_h) = \tfrac{1}{3}(e_0+e_1+e_2) \text{ for every } h.

Applying the Step-3 construction of the proof to this \(W\): take any linear projection \(\pi\) onto \(W\) (e.g. \(\pi(e_h)=\frac13(e_0+e_1+e_2)\) for each basis vector, extended linearly) and average; by \(G\)-symmetry of the regular representation the average is already \(G\)-invariant here, \(\tilde\pi(v) = \frac{1}{3}(v_0+v_1+v_2)(e_0+e_1+e_2)\) where \(v=\sum v_h e_h\). B

W' = \ker\tilde\pi = \Big\{\textstyle\sum_h c_h e_h : \sum_h c_h = 0\Big\}, \qquad \mathbb{C}G = W \oplus W'.

By Step 8 of the proof, \(W'=\ker\tilde\pi\) is \(G\)-invariant and complements \(W\); here it is the classical "augmentation ideal," visibly a 2-dimensional \(G\)-invariant subspace on which \(\rho\) further splits into the two nontrivial characters \(\omega,\omega^2\) of \(\mathbb{Z}/3\mathbb{Z}\) (\(\omega = e^{2\pi i/3}\)), giving the full decomposition \(\mathbb{C}G \cong \mathbf{1}\oplus\omega\oplus\omega^2\). B

\mathbb{C}[\mathbb{Z}/3\mathbb{Z}] \;\cong\; \mathbf{1} \oplus \omega \oplus \omega^2

Reading. The regular representation of a finite abelian group splits completely into its \(|G|\) one-dimensional characters, exactly as Maschke guarantees.

G = S_3, \quad V = \mathbb{C}^3 \text{ (permutation representation, permuting coordinates)}.

Standard setup: \(S_3\) acts on \(\mathbb{C}^3\) by permuting the standard basis \(e_1,e_2,e_3\). A

W = \mathbb{C}(e_1+e_2+e_3) \quad \text{is invariant (trivial subrepresentation)}.

Any permutation matrix fixes the all-ones vector, since it just permutes equal summands. A

\text{Since } |S_3| = 6 \text{ is invertible in } \mathbb{C}, \text{ Maschke guarantees an invariant complement } W'.

This is the direct hypothesis-check licensing use of the theorem: \(G=S_3\) finite, \(k=\mathbb{C}\) has characteristic \(0\). A

\text{Rather than reconstruct } \tilde\pi \text{ from scratch, use the averaged standard inner product on } \mathbb{C}^3 \text{ and take } W' = W^\perp.

This is the "unitary trick" corollary noted above: since the standard inner product on \(\mathbb{C}^3\) is already \(S_3\)-invariant (permutation matrices are orthogonal), \(W^{\perp}\) is automatically \(G\)-invariant — no further averaging of the inner product is needed here. B

W' = \{(x,y,z)\in\mathbb{C}^3 : x+y+z=0\}, \qquad \dim W' = 2.

Orthogonal complement of the line spanned by \((1,1,1)\) under the standard Hermitian form is exactly the hyperplane of coordinate sum zero. A

W' \text{ is irreducible: it is the 2-dimensional "standard representation" of } S_3, \text{ the unique irreducible of that dimension (since } 1^2+1^2+2^2=6=|S_3|\text{).}

Uses the corollary that \(\sum_i n_i^2 = |G|\) from the Artin–Wedderburn decomposition of \(\mathbb{C}S_3\), together with knowing \(S_3\) has exactly three irreducibles (trivial, sign, standard) since it has three conjugacy classes; \(1+1+4=6\) forces the last one to be irreducible of dimension \(2\), matching \(W'\). C

\mathbb{C}^3 \;\cong\; \mathbf{1} \;\oplus\; \text{std}_2 \quad \text{(trivial } \oplus \text{ the 2-dim irreducible of } S_3\text{)}

Reading. The permutation representation of \(S_3\) on \(\mathbb{C}^3\) is not irreducible, but Maschke guarantees — and the invariant inner product exhibits explicitly — a complementary invariant subspace, here the irreducible standard representation.

Problems

Verify directly (without citing Maschke) that the representation of \(\mathbb{Z}/2\mathbb{Z}\) on \(\mathbb{C}^2\) given by \(\rho(1)=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) is completely reducible, by exhibiting eigenvectors.
Solution
\(\rho(1)\) has eigenvalues \(\pm1\) with eigenvectors \(v_+=(1,1)\), \(v_-=(1,-1)\), since \(\rho(1)v_\pm = \pm v_\pm\). Each spans a \(G\)-invariant line (as \(\rho(0)=I\) trivially preserves everything and \(\rho(1)\) fixes/negates the eigenvector, staying in its own line). Hence \(\mathbb{C}^2 = \mathbb{C}v_+ \oplus \mathbb{C}v_-\), a direct sum of two \(1\)-dimensional (automatically irreducible) subrepresentations — the trivial and sign representations of \(\mathbb{Z}/2\mathbb{Z}\).
Let \(G=\mathbb{Z}/2\mathbb{Z}\) act on \(V=\mathbb{F}_2^2\) via \(\rho(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}\). Show explicitly that the averaging construction \(\tilde\pi = \frac{1}{|G|}\sum_g \rho(g)\pi\rho(g)^{-1}\) cannot be formed, and confirm no invariant complement to \(W=\mathbb{F}_2\cdot(1,0)\) exists.
Solution
\(|G|=2\), and in \(\mathbb{F}_2\), \(2=0\), so \(\frac{1}{|G|}=\frac12\) does not exist in \(\mathbb{F}_2\) — the formula is undefined, exactly as the theorem's hypothesis \(\operatorname{char}(k)\nmid|G|\) requires. Directly: \(W=\{(0,0),(1,0)\}\) is invariant since \(\rho(1)(1,0)=(1,0)\). The only other nonzero vectors are \((0,1)\) and \((1,1)\); \(\rho(1)(0,1)=(1,1)\ne(0,1)\), so the line through \((0,1)\) is not invariant, and it is the only candidate complementary line (as \(\mathbb{F}_2^2\) has exactly 3 nonzero vectors, one of which is in \(W\)). Hence no invariant complement exists.
Use Maschke's theorem to prove that any finite-dimensional representation \(\rho: G \to GL(V)\) of a finite group over \(\mathbb{C}\) is equivalent to a unitary representation (i.e. \(V\) carries a \(G\)-invariant Hermitian inner product).
Solution
Start with any Hermitian inner product \(\langle,\rangle_0\) on \(V\) (exists since \(V\) is a finite-dimensional \(\mathbb{C}\)-vector space). Define \(\langle u,v\rangle := \frac{1}{|G|}\sum_{g\in G}\langle \rho(g)u,\rho(g)v\rangle_0\). This is again Hermitian and positive-definite (a sum of positive-definite forms scaled by a positive real \(1/|G|\)), and for any \(h\in G\), \(\langle \rho(h)u,\rho(h)v\rangle = \frac1{|G|}\sum_g \langle \rho(gh)u,\rho(gh)v\rangle_0 = \langle u,v\rangle\) by reindexing \(g\mapsto gh\), a bijection of \(G\) — the same reindexing trick as Step 7 of the proof. Hence \(\langle,\rangle\) is \(G\)-invariant, so each \(\rho(g)\) is unitary with respect to it. (This gives an alternative proof of Maschke: given invariant \(W\), take \(W'=W^\perp\) under this invariant form, which is automatically \(G\)-invariant since unitary maps preserve orthogonal complements.)
Let \(G\) be a finite group and \(V\) a representation over \(\mathbb{C}\) with a \(G\)-invariant subspace \(W\). Suppose \(f:V\to W\) is any \(G\)-equivariant linear map with \(f|_W = \mathrm{id}_W\) (not assumed to arise from averaging). Show \(V = W \oplus \ker f\) directly, and explain why this does not need Maschke's averaging step.
Solution
Since \(f\) is already linear with \(f|_W=\mathrm{id}_W\) and image contained in \(W\) (as \(f\) maps into \(W\)), \(f\) is idempotent: for \(v\in V\), \(f(v)\in W\) so \(f(f(v))=f(v)\). Hence \(V=\mathrm{im}(f)\oplus\ker(f)=W\oplus\ker f\) by the standard idempotent decomposition. Since \(f\) is \(G\)-equivariant by hypothesis, \(\ker f\) is automatically \(G\)-invariant: if \(f(v)=0\) then \(f(\rho(h)v)=\rho(h)f(v)=0\). No averaging was needed because equivariance was assumed outright — Maschke's proof's only job is to manufacture such an \(f\) (namely \(\tilde\pi\)) starting from an arbitrary, non-equivariant projection \(\pi\), which is the hard part; this problem shows the "soft" part (equivariant idempotent \(\Rightarrow\) invariant splitting) is pure linear algebra.
Decompose the regular representation of \(G=\mathbb{Z}/4\mathbb{Z}\) over \(\mathbb{C}\) into irreducibles, and verify the dimension identity \(\sum_i n_i^2 = |G|\).
Solution
\(\mathbb{Z}/4\mathbb{Z}\) is abelian, so (by the general fact that abelian groups have only \(1\)-dimensional complex irreducibles, itself a consequence of Maschke plus Schur's lemma) every irreducible is a character \(\chi_k(n) = i^{kn}\) for \(k=0,1,2,3\), i.e. \(\chi_k(1) \in \{1,i,-1,-i\}\). Maschke's theorem guarantees the regular representation \(\mathbb{C}[\mathbb{Z}/4\mathbb{Z}]\) (dimension 4) splits completely, and since there are exactly \(4\) distinct characters, each must occur with multiplicity \(1\) (matching \(\dim \mathbb{C}G = |G|\) against \(\sum n_i^2\)): \(\mathbb{C}[\mathbb{Z}/4\mathbb{Z}] \cong \chi_0\oplus\chi_1\oplus\chi_2\oplus\chi_3\). Dimension check: \(1^2+1^2+1^2+1^2 = 4 = |\mathbb{Z}/4\mathbb{Z}|\), confirming the Artin–Wedderburn count.