[proofplan] Induct on $n$, the number of irreducible summands. The base case $n = 0$ is immediate: $V = 0$ and both sides are zero. The case $n = 1$ is exactly part 3 of Schur's lemma over an algebraically closed field, which gives $\dim \operatorname{Hom}_G(S, V) = 1$ if $S \cong V$ and $0$ otherwise — matching the count $|\{j : V_j \cong S\}|$. For the inductive step, split off the last summand to get $V = V' \oplus V_n$ with $V' = \bigoplus_{i=1}^{n-1} V_i$ a direct sum of $n - 1$ irreducibles. Hom Additivity converts $\dim \operatorname{Hom}_G(S, V)$ into $\dim \operatorname{Hom}_G(S, V') + \dim \operatorname{Hom}_G(S, V_n)$, the inductive hypothesis handles the first term, and the base case handles the second. Adding gives the result. [/proofplan]

[step:Set up the induction on the number of irreducible summands]Throughout, let $\mathbb{F}$ be algebraically closed and let $S$ be a fixed irreducible $G$-representation. We prove by induction on $n \geq 0$ the statement \begin{align*} P(n): \text{ for every decomposition } V = \bigoplus_{i=1}^n V_i \text{ into irreducible } G\text{-representations}, \\ \quad |\{j \in \{1, \ldots, n\} : V_j \cong S\}| = \dim_\mathbb{F} \operatorname{Hom}_G(S, V). \end{align*}[/step]

[guided]The natural inductive parameter is the number $n$ of irreducible summands. We fix the irreducible $S$ throughout — the statement is "for each $S$, the number of summands isomorphic to $S$ equals $\dim \operatorname{Hom}_G(S, V)$"; we prove this for one $S$ at a time, and the result holds for all $S$ since $S$ was arbitrary. The induction has two ingredients we must use carefully: 1. The base case ($n = 1$) is Schur's lemma, computing $\dim \operatorname{Hom}_G(S, V_1)$ for a single irreducible $V_1$. 2. The inductive step splits one summand off and applies Hom Additivity to the second factor of $V = V' \oplus V_n$.[/guided]

[step:Establish the base cases $n = 0$ and $n = 1$]**Case $n = 0$.** The decomposition $V = \bigoplus_{i \in \emptyset} V_i$ means $V = 0$. Then \begin{align*} \operatorname{Hom}_G(S, V) = \operatorname{Hom}_G(S, 0) = \{0\}, \end{align*} the only $G$-linear map into the zero space being the zero map, so $\dim_\mathbb{F} \operatorname{Hom}_G(S, V) = 0$. The set $\{j \in \emptyset : V_j \cong S\} = \emptyset$ has cardinality $0$. Both sides are $0$, so $P(0)$ holds. **Case $n = 1$.** The decomposition is $V = V_1$, so $V$ itself is irreducible. Apply part 3 of [Schur's Lemma](/theorems/2414): for irreducible $G$-representations $S, V_1$ over an algebraically closed field $\mathbb{F}$, \begin{align*} \dim_\mathbb{F} \operatorname{Hom}_G(S, V_1) = \begin{cases} 1 & \text{if } S \cong V_1, \\ 0 & \text{if } S \not\cong V_1. \end{cases} \end{align*} We verify the hypotheses of Schur 3: $\mathbb{F}$ is algebraically closed by global hypothesis; $S$ is irreducible by global choice; $V_1$ is irreducible because the decomposition $V = V_1$ has every summand irreducible by hypothesis. So Schur 3 applies. The right-hand side equals $|\{1\} \cap \{j : V_j \cong S\}|$: if $V_1 \cong S$ this is $1$, otherwise $0$. So $P(1)$ holds.[/step]

[guided]**$n = 0$.** An empty direct sum is the zero space, and $\operatorname{Hom}_G(S, 0)$ contains only the zero map (any $G$-linear map into $0$ sends everything to $0$). Both sides of the equation are $0$. **$n = 1$.** The decomposition is $V = V_1$ with $V_1$ irreducible, so $V$ is irreducible. We need $\dim \operatorname{Hom}_G(S, V_1)$. This is part 3 of Schur's lemma — the dimension of the Hom space between two irreducibles is $1$ if they are isomorphic, $0$ otherwise. Schur 3 requires (i) the base field is algebraically closed and (ii) both $S$ and $V_1$ are irreducible. We have (i) by global hypothesis; (ii) because $S$ was chosen irreducible at the start and the decomposition gives $V_1$ irreducible. So Schur 3 applies and gives the count. The right-hand side $|\{j : V_j \cong S\}|$ is $1$ when $V_1 \cong S$, and $0$ otherwise — matching Schur 3.[/guided]

[step:Carry out the inductive step using Hom Additivity]Assume $P(n-1)$ holds for some $n \geq 2$ and let $V = \bigoplus_{i=1}^n V_i$ be a decomposition into irreducibles. Set \begin{align*} V' := \bigoplus_{i=1}^{n-1} V_i, \end{align*} a $G$-representation with a decomposition into $n - 1$ irreducibles. Then $V = V' \oplus V_n$. Apply [Hom Additivity](/theorems/2419) with $V \rightsquigarrow S$, $V_1 \rightsquigarrow V'$, $V_2 \rightsquigarrow V_n$ (using the hypothesis that all $V', V_n$ are $G$-representations over the same field $\mathbb{F}$, which they are by construction). Part (1) of Hom Additivity gives an $\mathbb{F}$-linear isomorphism \begin{align*} \operatorname{Hom}_G(S, V' \oplus V_n) \cong \operatorname{Hom}_G(S, V') \oplus \operatorname{Hom}_G(S, V_n). \end{align*} Taking $\mathbb{F}$-dimensions on both sides (using $\dim(A \oplus B) = \dim A + \dim B$ for $\mathbb{F}$-vector spaces $A, B$): \begin{align*} \dim_\mathbb{F} \operatorname{Hom}_G(S, V) = \dim_\mathbb{F} \operatorname{Hom}_G(S, V') + \dim_\mathbb{F} \operatorname{Hom}_G(S, V_n). \end{align*} By the inductive hypothesis $P(n-1)$ applied to the decomposition $V' = \bigoplus_{i=1}^{n-1} V_i$, \begin{align*} \dim_\mathbb{F} \operatorname{Hom}_G(S, V') = |\{j \in \{1, \ldots, n-1\} : V_j \cong S\}|. \end{align*} By the base case $P(1)$ applied to the irreducible $V_n$, \begin{align*} \dim_\mathbb{F} \operatorname{Hom}_G(S, V_n) = \begin{cases} 1 & V_n \cong S, \\ 0 & V_n \not\cong S, \end{cases} \end{align*} which equals $|\{n\} \cap \{j : V_j \cong S\}|$. Adding, \begin{align*} \dim_\mathbb{F} \operatorname{Hom}_G(S, V) &= |\{j \in \{1, \ldots, n-1\} : V_j \cong S\}| + |\{n\} \cap \{j : V_j \cong S\}| \\ &= |\{j \in \{1, \ldots, n\} : V_j \cong S\}|, \end{align*} the second equality being the disjoint partition $\{1, \ldots, n\} = \{1, \ldots, n-1\} \sqcup \{n\}$. This is $P(n)$.[/step]

[guided]**Splitting off the last summand.** Group the first $n - 1$ irreducibles into $V' := \bigoplus_{i=1}^{n-1} V_i$ and write $V = V' \oplus V_n$. The induction hypothesis $P(n-1)$ tells us about $\operatorname{Hom}_G(S, V')$ — it counts the $S$-isomorphic summands among $V_1, \ldots, V_{n-1}$. **Why we use Hom Additivity here.** Hom Additivity (part 1) splits Hom out of $S$ into a direct sum across the codomain: \begin{align*} \operatorname{Hom}_G(S, V' \oplus V_n) \cong \operatorname{Hom}_G(S, V') \oplus \operatorname{Hom}_G(S, V_n). \end{align*} Hom Additivity has no hypothesis beyond "$V, V_1, V_2$ are $G$-representations over $\mathbb{F}$", which is satisfied here ($S, V', V_n$ are all $G$-representations over $\mathbb{F}$ by construction). **Taking dimensions.** $\mathbb{F}$-dimensions add along direct sums: $\dim(A \oplus B) = \dim A + \dim B$. So we obtain a numerical identity: \begin{align*} \dim \operatorname{Hom}_G(S, V) = \dim \operatorname{Hom}_G(S, V') + \dim \operatorname{Hom}_G(S, V_n). \end{align*} **Applying the inductive hypothesis and the base case.** The first term is by induction the count of $V_i \cong S$ among $i = 1, \ldots, n-1$. The second term is by the base case ($V_n$ alone is irreducible) the indicator of $V_n \cong S$. Adding gives the count among $i = 1, \ldots, n$. This is $P(n)$, completing the induction.[/guided]

[step:Conclude the multiplicity formula]By induction $P(n)$ holds for every $n \geq 0$. Hence for any decomposition $V = \bigoplus_{i=1}^n V_i$ into irreducible $G$-representations and any irreducible $G$-representation $S$, \begin{align*} |\{j \in \{1, \ldots, n\} : V_j \cong S\}| = \dim_\mathbb{F} \operatorname{Hom}_G(S, V). \end{align*} This is the multiplicity formula stated in the theorem.[/step]

[guided]The induction closes: $P(0)$ and $P(1)$ are the base cases (the latter being Schur 3), and $P(n-1) \implies P(n)$ via Hom Additivity. So $P(n)$ for all $n \geq 0$. The formula has a clean reading: $\dim \operatorname{Hom}_G(S, V)$ is a basis-free way to extract the multiplicity of $S$ in any decomposition of $V$. In particular, the multiplicity does **not** depend on the chosen decomposition — different decompositions of $V$ into irreducibles must agree in how many copies of each $S$ they use, recovering the uniqueness part of complete reducibility (in this guise, a Krull-Schmidt-type statement).[/guided]