Step 1: Construct Specht modules

For each partition $\lambda \vdash n$, construct $S^\lambda$ using Young symmetrizers:

• Choose a Young tableau $T$ of shape $\lambda$
• Define row group $R_T$ and column group $C_T$
• Form $c_T = b_T a_T$ where $a_T = \sum_{r \in R_T} r$ and $b_T = \sum_{c \in C_T} \text{sgn}(c) c$
• Set $S^\lambda = \mathbb{C}[S_n] c_T$

Step 2: Show $S^\lambda$ is irreducible

The key is to show $S^\lambda$ has no proper non-zero $S_n$-submodules. This uses:

• The standard polytabloid basis $\{e_T : T \text{ standard}\}$
• Young's orthogonal form: an inner product making $S^\lambda$ orthogonal to other Specht modules
• The straightening algorithm: any element can be written in terms of standard polytabloids

Step 3: Show $S^\lambda$ are pairwise non-isomorphic

For $\lambda \neq \mu$, we have $S^\lambda \not\cong S^\mu$. This follows from:

• Different dimensions: $f^\lambda \neq f^\mu$ for distinct $\lambda, \mu$
• Different character values: The characters are distinct functions on $S_n$
• Branching rules: Restrictions to $S_{n-1}$ distinguish representations

Step 4: Count dimensions

We must verify: $\sum_{\lambda \vdash n} (f^\lambda)^2 = n!$

This follows from decomposing the regular representation: \mathbb{C}[S_n] \cong \bigoplus_{\lambda \vdash n} S^\lambda^{\oplus f^\lambda}

Taking dimensions: $n! = \sum (f^\lambda \cdot f^\lambda) = \sum (f^\lambda)^2$.

Step 5: Completeness

Since the dimensions account for all of $\mathbb{C}[S_n]$ and we've shown irreducibility, the Specht modules form a complete set of irreducibles.