[proofplan]
We first use [Maschke's theorem](/theorems/2409) to know that the left regular module $\mathbb C[S_n]$ is a direct sum of irreducible complex $S_n$-modules. The irreducible modules are precisely the Specht modules $S^\lambda$ indexed by partitions $\lambda \vdash n$, so it remains only to compute their multiplicities. The multiplicity of an irreducible module $V$ in the regular module is $\dim_{\mathbb C} V$; we prove this directly by identifying $\operatorname{Hom}_{\mathbb C[S_n]}(V,\mathbb C[S_n])$ with the dual [vector space](/page/Vector%20Space) $V^*$. Substituting $V=S^\lambda$ gives the claimed decomposition.
[/proofplan]
[step:Apply Maschke's theorem and Specht classification to reduce to multiplicities]
Set $G:=S_n$ and set $A:=\mathbb C[G]$. The algebra $A$ is a left $A$-module by left multiplication. Since $G$ is finite and $\operatorname{char}(\mathbb C)=0$ does not divide $|G|$, [citetheorem:8439] implies that every finite-dimensional complex $G$-module is completely reducible. In particular, $A$ is a finite-dimensional semisimple left $A$-module.
By [citetheorem:8440], the irreducible finite-dimensional complex $G$-modules are, up to isomorphism, precisely the Specht modules $S^\lambda$ with $\lambda \vdash n$, and no two of these are isomorphic. Therefore there exist unique integers $m_\lambda \ge 0$ such that
\begin{align*}
A \cong \bigoplus_{\lambda \vdash n} (S^\lambda)^{\oplus m_\lambda}
\end{align*}
as left $A$-modules. It remains to prove that $m_\lambda=\dim_{\mathbb C}S^\lambda$ for every partition $\lambda \vdash n$.
[guided]
We introduce the abbreviations $G:=S_n$ and $A:=\mathbb C[G]$ so that the proof is a statement about the regular module of a finite group algebra. The left regular module is $A$ itself, with the action
\begin{align*}
A \times A &\to A
\end{align*}
\begin{align*}
(a,b) &\mapsto ab.
\end{align*}
The first structural input is Maschke's theorem. Its hypotheses are satisfied because $G=S_n$ is finite and the base field is $\mathbb C$, whose characteristic is $0$, so $\operatorname{char}(\mathbb C)$ does not divide $|G|$. Thus [citetheorem:8439] tells us that every finite-dimensional complex $G$-module is completely reducible. Since $A=\mathbb C[G]$ has basis $G$, it has complex dimension $|G|$, so it is finite-dimensional. Hence $A$ decomposes as a direct sum of irreducible left $A$-modules.
The second structural input identifies which irreducibles can occur. By [citetheorem:8440], the irreducible complex $S_n$-modules are exactly the Specht modules $S^\lambda$ indexed by partitions $\lambda \vdash n$, with no repetitions up to isomorphism. Therefore the regular module has a decomposition
\begin{align*}
A \cong \bigoplus_{\lambda \vdash n} (S^\lambda)^{\oplus m_\lambda}
\end{align*}
for some integers $m_\lambda \ge 0$. The only remaining task is numerical: compute each multiplicity $m_\lambda$.
[/guided]
[/step]
[step:Compute the multiplicity of an irreducible module in the regular module]
Let $V$ be an irreducible finite-dimensional left $A$-module. We prove that the multiplicity of $V$ in $A$ is $\dim_{\mathbb C}V$.
Define a complex-[linear map](/page/Linear%20Map)
\begin{align*}
\Phi:\operatorname{Hom}_A(V,A) &\to V^*
\end{align*}
\begin{align*}
T &\mapsto \varepsilon \circ T,
\end{align*}
where $\varepsilon:A\to\mathbb C$ is the coefficient-of-identity map
\begin{align*}
\varepsilon\left(\sum_{g\in G} a_g g\right)=a_{1_G}.
\end{align*}
For each $\ell\in V^*$, define
\begin{align*}
T_\ell:V &\to A
\end{align*}
\begin{align*}
v &\mapsto \sum_{g\in G} \ell(g^{-1}v)g.
\end{align*}
For $h\in G$ and $v\in V$,
\begin{align*}
T_\ell(hv)=\sum_{g\in G}\ell(g^{-1}hv)g.
\end{align*}
Using the substitution $k=h^{-1}g$, equivalently $g=hk$, this becomes
\begin{align*}
T_\ell(hv)=\sum_{k\in G}\ell(k^{-1}v)hk=hT_\ell(v).
\end{align*}
Thus $T_\ell$ is $A$-linear. Moreover,
\begin{align*}
(\varepsilon\circ T_\ell)(v)=\ell(v),
\end{align*}
so $\Phi(T_\ell)=\ell$.
Conversely, if $T\in\operatorname{Hom}_A(V,A)$ and $\ell=\varepsilon\circ T$, write
\begin{align*}
T(v)=\sum_{g\in G} a_g(v)g
\end{align*}
for uniquely determined scalars $a_g(v)\in\mathbb C$. Since $T$ is $A$-linear,
\begin{align*}
g^{-1}T(v)=T(g^{-1}v).
\end{align*}
The coefficient of $1_G$ in the left-hand side is $a_g(v)$, while the coefficient of $1_G$ in the right-hand side is $\ell(g^{-1}v)$. Hence $a_g(v)=\ell(g^{-1}v)$ for every $g\in G$, so $T=T_\ell$. Therefore $\Phi$ is an isomorphism, and
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_A(V,A)=\dim_{\mathbb C}V^*=\dim_{\mathbb C}V.
\end{align*}
If
\begin{align*}
A \cong \bigoplus_{\mu \vdash n} (S^\mu)^{\oplus m_\mu},
\end{align*}
then applying $\operatorname{Hom}_A(V,-)$ and using [Schur's lemma](/theorems/2414) gives
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_A(V,A)=m_\lambda
\end{align*}
when $V\cong S^\lambda$. Hence $m_\lambda=\dim_{\mathbb C}S^\lambda$.
[/step]
[step:Substitute the Specht dimensions and conclude]
For each partition $\lambda \vdash n$, the theorem defines
\begin{align*}
f_\lambda:=\dim_{\mathbb C}S^\lambda.
\end{align*}
The previous step gives $m_\lambda=\dim_{\mathbb C}S^\lambda=f_\lambda$. Therefore
\begin{align*}
\mathbb C[S_n]\cong \bigoplus_{\lambda\vdash n}(S^\lambda)^{\oplus f_\lambda}
\end{align*}
as left $\mathbb C[S_n]$-modules, which is the desired decomposition.
[/step]
