[proofplan] We argue by strong induction on $\dim V$. The base case $\dim V = 0$ is the empty direct sum. For the inductive step, if $V$ is irreducible then we are done; otherwise there exists a proper non-zero $G$-invariant subspace $W \subsetneq V$. Maschke's theorem produces a $G$-invariant complement $W'$ with $V = W \oplus W'$ as $G$-spaces. Since $W$ and $W'$ are non-zero, $\dim W < \dim V$ and $\dim W' < \dim V$, so the inductive hypothesis applies to both summands and decomposes them into irreducibles. Concatenating the two decompositions gives a decomposition of $V$ into irreducibles. [/proofplan] [step:Set up strong induction on $\dim V$ and dispose of the zero-dimensional base case] Let $G$ be a finite group and $\mathbb{F}$ a field of characteristic $0$. We prove by strong induction on $n = \dim_\mathbb{F} V$ the statement: > $P(n)$: every $n$-dimensional $G$-representation $V$ over $\mathbb{F}$ decomposes as a direct sum of irreducible $G$-subrepresentations. **Base case $n = 0$.** If $\dim V = 0$, then $V = \{0\}$ and the empty direct sum (a sum indexed by the empty set, conventionally equal to $\{0\}$) is the required decomposition. We allow $n = 1$ to fall into the inductive step below: a one-dimensional representation has no proper non-zero subspaces, so it is irreducible by definition, and the inductive step handles this case directly. [/step] [step:Treat the irreducible case as a one-summand decomposition] Fix $n \geq 1$ and assume $P(m)$ holds for all $0 \leq m < n$. Let $V$ be an $n$-dimensional $G$-representation. There are two mutually exclusive cases. **Case 1: $V$ is irreducible.** Then $V$ itself is the required decomposition: $V = V$ as a single-term direct sum of irreducibles. **Case 2: $V$ is not irreducible.** Then by definition there exists a $G$-invariant subspace $W \leq V$ with $W \neq \{0\}$ and $W \neq V$. The next step splits off a $G$-invariant complement. [/step] [step:Extract a $G$-invariant complement to $W$ via Maschke's theorem] Suppose we are in Case 2 of the previous step, so $W \leq V$ is a proper non-zero $G$-invariant subspace. We invoke [Maschke's Theorem](/theorems/2409) to produce a complement. **Verification of hypotheses.** Maschke's theorem requires (i) $G$ is finite — given by hypothesis; (ii) $\operatorname{char}\mathbb{F} \nmid |G|$. Since $\operatorname{char}\mathbb{F} = 0$ by hypothesis, no positive integer (and in particular no positive divisor of $|G|$) is zero in $\mathbb{F}$, so $\operatorname{char}\mathbb{F} \nmid |G|$ holds vacuously. **Conclusion.** Maschke's theorem provides a $G$-invariant subspace $W' \leq V$ such that $V = W \oplus W'$ as $G$-representations. [guided] We have a proper non-zero $G$-invariant subspace $W \leq V$, and we need to "split it off" — find another $G$-invariant subspace $W'$ such that $V = W \oplus W'$. This is exactly what [Maschke's Theorem](/theorems/2409) provides. Why is Maschke's theorem applicable? It has two hypotheses, and we verify each: 1. **$G$ is finite.** This is one of the hypotheses of the theorem we are currently proving, so it is automatic. 2. **The characteristic of $\mathbb{F}$ does not divide $|G|$.** This is where the hypothesis $\operatorname{char}\mathbb{F} = 0$ enters. In characteristic $0$, no positive integer is zero in $\mathbb{F}$, so in particular $|G|$ is non-zero in $\mathbb{F}$ and is therefore not divisible (in the appropriate sense) by the characteristic. Equivalently: $|G|$ is invertible in $\mathbb{F}$, which is the form in which the hypothesis is actually consumed inside the proof of Maschke's theorem (Maschke's argument averages over $G$ and divides by $|G|$). Why is the characteristic-$0$ hypothesis really needed? In characteristic $p$ dividing $|G|$, the result genuinely fails: e.g., over $\mathbb{F}_p$ the representation of $C_p$ on $\mathbb{F}_p^2$ by upper-triangular unipotent matrices has the unique subspace $\langle e_1 \rangle$ as a $G$-invariant proper subspace, but no $G$-invariant complement exists. So characteristic $0$ (or, more generally, $\operatorname{char}\mathbb{F} \nmid |G|$) is essential. With both hypotheses verified, Maschke's theorem gives a $G$-invariant subspace $W' \leq V$ with $V = W \oplus W'$ as $G$-representations. [/guided] [/step] [step:Apply the inductive hypothesis to $W$ and $W'$ and concatenate] We have $V = W \oplus W'$ with both $W$ and $W'$ non-zero $G$-invariant subspaces of $V$. Since $V = W \oplus W'$, we have $\dim V = \dim W + \dim W'$. Both summands are positive (as $W \neq 0$ and $W' \neq 0$), so \begin{align*} \dim W < \dim V = n \qquad \text{and} \qquad \dim W' < \dim V = n. \end{align*} By the strong inductive hypothesis $P(\dim W)$ and $P(\dim W')$, there exist decompositions \begin{align*} W &= U_1 \oplus \cdots \oplus U_p, \\ W' &= U'_1 \oplus \cdots \oplus U'_q, \end{align*} where each $U_i$ and each $U'_j$ is an irreducible $G$-subrepresentation. Concatenating: \begin{align*} V = W \oplus W' = U_1 \oplus \cdots \oplus U_p \oplus U'_1 \oplus \cdots \oplus U'_q, \end{align*} which is a decomposition of $V$ as a direct sum of irreducible $G$-subrepresentations. This establishes $P(n)$ and completes the induction. [guided] We are at the heart of the inductive argument. Maschke's theorem has just split $V$ into two pieces $V = W \oplus W'$, both of which are $G$-invariant and both of which are non-zero (we chose $W$ to be proper and non-zero, so $W'$ must also be non-zero — otherwise $V = W$, contradicting properness). The crucial observation is that **both pieces have strictly smaller dimension than $V$**: \begin{align*} \dim V = \dim W + \dim W', \end{align*} and since both $\dim W \geq 1$ and $\dim W' \geq 1$, we get $\dim W \leq n - 1 < n$ and $\dim W' \leq n - 1 < n$. Why does this matter? Because we are doing **strong induction**: our inductive hypothesis says that every representation of dimension strictly less than $n$ is completely reducible. So we can apply it to both $W$ and $W'$, obtaining \begin{align*} W &= U_1 \oplus \cdots \oplus U_p \qquad \text{(each } U_i \text{ irreducible),} \\ W' &= U'_1 \oplus \cdots \oplus U'_q \qquad \text{(each } U'_j \text{ irreducible).} \end{align*} Now we simply concatenate these two decompositions, using the associativity of direct sum: \begin{align*} V = W \oplus W' = (U_1 \oplus \cdots \oplus U_p) \oplus (U'_1 \oplus \cdots \oplus U'_q) = U_1 \oplus \cdots \oplus U_p \oplus U'_1 \oplus \cdots \oplus U'_q. \end{align*} Each summand on the right is an irreducible $G$-subrepresentation, so this is the required decomposition of $V$. The induction is complete. A subtlety worth flagging: the induction terminates **because each splitting strictly decreases dimension**. Without Maschke's theorem we could have produced a $G$-invariant subspace $W$, but no guarantee that the remaining piece $V/W$ (or any concrete complement) was again a $G$-representation. Maschke's theorem turns the abstract submodule structure into a literal direct-sum decomposition, which is what makes the induction work. [/guided] [/step]