[proofplan] We compute the characters of the Specht modules through the Frobenius characteristic map. The Specht character of shape $\lambda$ has Frobenius characteristic $s_\lambda$, and the Frobenius characteristic identifies the usual character [inner product](/page/Inner%20Product) on class functions with the Hall inner product on symmetric functions. Since the Schur functions are orthonormal, the Specht characters have inner product $1$ when the indexing partitions are equal and $0$ when they are distinct. The character criterion then gives irreducibility and pairwise non-isomorphism, and the standard count of irreducible complex representations by conjugacy classes gives completeness because conjugacy classes of $S_n$ are indexed by partitions of $n$. [/proofplan] [step:Declare the character and Frobenius characteristic notation] For each partition $\lambda \vdash n$, let \begin{align*} \chi_\lambda: S_n \to \mathbb C \end{align*} be the character of the finite-dimensional $\mathbb C[S_n]$-module $S^\lambda$. Let $\operatorname{CF}(S_n)$ denote the complex [vector space](/page/Vector%20Space) of class functions on $S_n$, equipped with the character inner product \begin{align*} \langle \alpha,\beta\rangle_{S_n}:=\frac{1}{n!}\sum_{w\in S_n}\alpha(w)\overline{\beta(w)} \end{align*} for class functions $\alpha,\beta:S_n\to\mathbb C$. Let $\Lambda^n$ denote the complex vector space of homogeneous symmetric functions of degree $n$, equipped with the Hall inner product $\langle\cdot,\cdot\rangle_{\mathrm{Hall}}$. Let \begin{align*} \operatorname{ch}:\operatorname{CF}(S_n)\to \Lambda^n \end{align*} be the Frobenius characteristic map. We use the standard Frobenius characteristic isometry: \begin{align*} \langle \alpha,\beta\rangle_{S_n}=\langle \operatorname{ch}(\alpha),\operatorname{ch}(\beta)\rangle_{\mathrm{Hall}} \end{align*} for all $\alpha,\beta\in\operatorname{CF}(S_n)$, and the standard Specht character calculation \begin{align*} \operatorname{ch}(\chi_\lambda)=s_\lambda \end{align*} for every $\lambda\vdash n$; this is the Frobenius characteristic form of Young's rule for Specht characters. Also, the Schur functions $\{s_\lambda:\lambda\vdash n\}$ form an [orthonormal basis](/page/Orthonormal%20Basis) of $\Lambda^n$ for the Hall inner product. These three standard symmetric-function results are being used as prerequisites here: Frobenius characteristic isometry, Specht character has Frobenius characteristic $s_\lambda$, and Schur orthonormality. [/step] [step:Compute the inner products of the Specht characters] Let $\lambda,\mu\vdash n$. Define the Kronecker delta symbol $\delta_{\lambda\mu}\in\{0,1\}$ by setting $\delta_{\lambda\mu}=1$ if $\lambda=\mu$ and $\delta_{\lambda\mu}=0$ if $\lambda\ne\mu$. Applying the Frobenius characteristic isometry to the two class functions $\chi_\lambda$ and $\chi_\mu$ gives \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\langle \operatorname{ch}(\chi_\lambda),\operatorname{ch}(\chi_\mu)\rangle_{\mathrm{Hall}}. \end{align*} Using the Specht character calculation $\operatorname{ch}(\chi_\lambda)=s_\lambda$ and $\operatorname{ch}(\chi_\mu)=s_\mu$, this becomes \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\langle s_\lambda,s_\mu\rangle_{\mathrm{Hall}}. \end{align*} By Schur orthonormality, \begin{align*} \langle s_\lambda,s_\mu\rangle_{\mathrm{Hall}}=\delta_{\lambda\mu}. \end{align*} Therefore \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\delta_{\lambda\mu}. \end{align*} [guided] The point of introducing the Frobenius characteristic is that it turns a representation-theoretic character computation into an orthonormality computation for Schur functions. Fix two partitions $\lambda,\mu\vdash n$, and define the Kronecker delta symbol $\delta_{\lambda\mu}\in\{0,1\}$ by setting $\delta_{\lambda\mu}=1$ if $\lambda=\mu$ and $\delta_{\lambda\mu}=0$ if $\lambda\ne\mu$. The characters \begin{align*} \chi_\lambda:S_n\to\mathbb C \end{align*} and \begin{align*} \chi_\mu:S_n\to\mathbb C \end{align*} are class functions because [characters are constant on conjugacy classes](/theorems/5008). We apply the Frobenius characteristic isometry to these two class functions. Its hypotheses are satisfied because both $\chi_\lambda$ and $\chi_\mu$ lie in $\operatorname{CF}(S_n)$. Hence \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\langle \operatorname{ch}(\chi_\lambda),\operatorname{ch}(\chi_\mu)\rangle_{\mathrm{Hall}}. \end{align*} Now the Specht character calculation identifies the two Frobenius characteristics: \begin{align*} \operatorname{ch}(\chi_\lambda)=s_\lambda \end{align*} and \begin{align*} \operatorname{ch}(\chi_\mu)=s_\mu. \end{align*} Substituting these identities into the previous display gives \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\langle s_\lambda,s_\mu\rangle_{\mathrm{Hall}}. \end{align*} The Schur functions of degree $n$ are orthonormal for the Hall inner product. Since $\lambda$ and $\mu$ are both partitions of $n$, this gives \begin{align*} \langle s_\lambda,s_\mu\rangle_{\mathrm{Hall}}=\delta_{\lambda\mu}. \end{align*} Combining the equalities yields \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=\delta_{\lambda\mu}. \end{align*} This is the decisive calculation: the diagonal value $1$ will force irreducibility, and the off-diagonal value $0$ will force distinct irreducibles not to be isomorphic. [/guided] [/step] [step:Use character orthonormality to prove irreducibility and pairwise non-isomorphism] Since $\mathbb C$ has characteristic $0$, the characteristic of $\mathbb C$ does not divide $|S_n|=n!$. By [citetheorem:8439], every finite-dimensional $\mathbb C[S_n]$-module is completely reducible. We now use the standard character inner product criterion for finite-dimensional complex representations of a finite group: a nonzero representation with character $\chi$ is irreducible if and only if $\langle\chi,\chi\rangle=1$, and two irreducible representations with characters $\chi$ and $\psi$ are isomorphic if and only if $\langle\chi,\psi\rangle=1$. This is a standard character-theoretic prerequisite not yet separately cited in the wiki. Taking $\mu=\lambda$ in the previous step gives \begin{align*} \langle \chi_\lambda,\chi_\lambda\rangle_{S_n}=1. \end{align*} Thus $S^\lambda$ is nonzero and irreducible. If $\lambda\ne\mu$, then \begin{align*} \langle \chi_\lambda,\chi_\mu\rangle_{S_n}=0. \end{align*} Since $S^\lambda$ and $S^\mu$ are irreducible, this implies $S^\lambda\not\cong S^\mu$ as $\mathbb C[S_n]$-modules. [/step] [step:Count the irreducibles to prove completeness] The conjugacy classes of $S_n$ are indexed by cycle type, and cycle types are exactly partitions of $n$. Hence the number of conjugacy classes of $S_n$ is \begin{align*} |\{\lambda:\lambda\vdash n\}|. \end{align*} For a finite group over $\mathbb C$, the number of isomorphism classes of irreducible finite-dimensional complex representations equals the number of conjugacy classes of the group; this is a standard finite group character theory theorem not yet separately cited in the wiki. Therefore $S_n$ has exactly \begin{align*} |\{\lambda:\lambda\vdash n\}| \end{align*} irreducible complex representations up to isomorphism. The preceding step produced the same number of pairwise non-isomorphic irreducible $\mathbb C[S_n]$-modules, namely one $S^\lambda$ for each partition $\lambda\vdash n$. Therefore every irreducible finite-dimensional $\mathbb C[S_n]$-module is isomorphic to exactly one Specht module $S^\lambda$. This proves irreducibility, pairwise non-isomorphism, and completeness of the family. [/step]