[proofplan] Fix a tableau $t$ of shape $\lambda$ and compare two standard models: the left ideal $\mathbb C[S_n]a_t$ and the tabloid permutation module $M^\lambda$. The map $xa_t\mapsto x\{t\}$, with $x$ ranging over left coset representatives for the row stabilizer, is shown to be an $S_n$-module isomorphism. Restricting this isomorphism to the left ideal $\mathbb C[S_n]c_t=\mathbb C[S_n]b_ta_t$ sends each generator $xc_t$ to the corresponding translate $xe_t$ of the polytabloid. Since every polytabloid of shape $\lambda$ is such a translate, the image is exactly the polytabloid span, and injectivity follows from the ambient isomorphism. [/proofplan] [step:Identify the permutation module with the row-symmetrizer left ideal] Fix a $\lambda$-tableau $t$. Let $R_t\le S_n$ denote the subgroup preserving each row of $t$ setwise, and define the row symmetrizer \begin{align*} a_t=\sum_{r\in R_t}r\in\mathbb C[S_n]. \end{align*} Let $X\subset S_n$ be a set of representatives for the left cosets of $R_t$ in $S_n$. Since $r a_t=a_t$ for every $r\in R_t$, the elements $xa_t$ with $x\in X$ span $\mathbb C[S_n]a_t$. They are linearly independent because their supports in the group basis of $\mathbb C[S_n]$ are the disjoint subsets $xR_t$. Hence \begin{align*} \{xa_t:x\in X\} \end{align*} is a $\mathbb C$-basis of $\mathbb C[S_n]a_t$. The tabloids $\{xt\}$ with $x\in X$ are exactly the distinct $\lambda$-tabloids, because two tableaux $xt$ and $yt$ determine the same tabloid precisely when $y^{-1}x\in R_t$, equivalently when $xR_t=yR_t$. Thus \begin{align*} \{\{xt\}:x\in X\} \end{align*} is the tabloid basis of $M^\lambda$. [guided] The purpose of this step is to put the group-algebra model and the tabloid model on the same indexing set. We fix a $\lambda$-tableau $t$ and define \begin{align*} a_t=\sum_{r\in R_t}r\in\mathbb C[S_n], \end{align*} where $R_t\le S_n$ is the subgroup whose elements preserve every row of $t$ setwise. Choose a subset $X\subset S_n$ containing exactly one representative of each left coset of $R_t$. For every $r_0\in R_t$ we have \begin{align*} r_0a_t=\sum_{r\in R_t}r_0r=a_t, \end{align*} because left multiplication by $r_0$ permutes the elements of $R_t$. Therefore, if $x\in S_n$ and $x=yr_0$ with $y\in X$ and $r_0\in R_t$, then \begin{align*} xa_t=yr_0a_t=ya_t. \end{align*} This proves that the elements $xa_t$ with $x\in X$ span $\mathbb C[S_n]a_t$. They are also linearly independent. Indeed, in the group basis of $\mathbb C[S_n]$, the support of $xa_t$ is the set $xR_t$. The sets $xR_t$ for $x\in X$ are pairwise disjoint left cosets. Thus a linear relation among the $xa_t$ would give zero coefficients on disjoint subsets of the group basis, so all coefficients must vanish. Hence $\{xa_t:x\in X\}$ is a basis of $\mathbb C[S_n]a_t$. Now compare with tabloids. The tabloid $\{xt\}$ remembers the rows of $xt$ but forgets the order inside each row. Thus $\{xt\}=\{yt\}$ holds exactly when $y^{-1}x$ rearranges entries only inside the rows of $t$, that is, exactly when $y^{-1}x\in R_t$. This is equivalent to $xR_t=yR_t$. Since $X$ contains one representative of each left coset, the tabloids $\{\{xt\}:x\in X\}$ are precisely the distinct $\lambda$-tabloids, and therefore form the standard basis of $M^\lambda$. [/guided] [/step] [step:Construct the equivariant isomorphism from $\mathbb C[S_n]a_t$ to $M^\lambda$] Define a $\mathbb C$-[linear map](/page/Linear%20Map) \begin{align*} \Phi:\mathbb C[S_n]a_t\to M^\lambda \end{align*} by prescribing its values on the basis from the previous step: \begin{align*} \Phi(xa_t)=\{xt\}\qquad\text{for }x\in X. \end{align*} Since this sends a basis of $\mathbb C[S_n]a_t$ bijectively to the tabloid basis of $M^\lambda$, the map $\Phi$ is a complex vector-space isomorphism. We verify $S_n$-equivariance. Let $\sigma\in S_n$ and $x\in X$. Choose the unique element $y\in X$ and an element $r\in R_t$ such that $\sigma x=yr$. Then \begin{align*} \Phi(\sigma xa_t)=\Phi(yra_t)=\Phi(ya_t)=\{yt\}. \end{align*} Since $r$ only permutes entries within rows of $t$, $\{yrt\}=\{yt\}$, and hence \begin{align*} \sigma\Phi(xa_t)=\sigma\{xt\}=\{\sigma xt\}=\{yrt\}=\{yt\}. \end{align*} Thus $\Phi(\sigma xa_t)=\sigma\Phi(xa_t)$ on the basis $\{xa_t:x\in X\}$, so $\Phi$ is an isomorphism of $\mathbb C[S_n]$-modules. [/step] [step:Compute the image of the Young symmetrizer left ideal] Let $C_t\le S_n$ be the subgroup preserving each column of $t$ setwise, and define \begin{align*} b_t=\sum_{c\in C_t}\operatorname{sgn}(c)c\in\mathbb C[S_n], \end{align*} with the convention \begin{align*} c_t=b_ta_t. \end{align*} The group-algebra Specht module is \begin{align*} S^\lambda=\mathbb C[S_n]c_t\subset \mathbb C[S_n]a_t. \end{align*} For each $w\in S_n$, linearity of $\Phi$ gives \begin{align*} \Phi(wc_t)=\Phi(wb_ta_t). \end{align*} Expanding $b_t$, \begin{align*} \Phi(wb_ta_t)=\sum_{c\in C_t}\operatorname{sgn}(c)\Phi(wca_t). \end{align*} Equivariance gives, for every $g\in S_n$, \begin{align*} \Phi(ga_t)=g\Phi(a_t)=g\{t\}=\{gt\}. \end{align*} Applying this consequence with $g=wc$ for each $c\in C_t$, \begin{align*} \sum_{c\in C_t}\operatorname{sgn}(c)\Phi(wca_t)=\sum_{c\in C_t}\operatorname{sgn}(c)\{wct\}=w e_t. \end{align*} Therefore \begin{align*} \Phi(\mathbb C[S_n]c_t)=\operatorname{span}_{\mathbb C}\{we_t:w\in S_n\}. \end{align*} [/step] [step:Identify the translated polytabloids with all polytabloids of shape $\lambda$] Let $u$ be any $\lambda$-tableau. Since $S_n$ acts transitively on the tableaux of shape $\lambda$ with entries $\{1,\dots,n\}$, there exists $w\in S_n$ such that $u=wt$. We show that $we_t=e_u$. The column stabilizer of $u=wt$ is \begin{align*} C_u=wC_tw^{-1}. \end{align*} Using invariance of the sign character under conjugation, \begin{align*} e_u=\sum_{d\in C_u}\operatorname{sgn}(d)\{du\}. \end{align*} Writing $d=wcw^{-1}$ with $c\in C_t$, this becomes \begin{align*} e_u=\sum_{c\in C_t}\operatorname{sgn}(c)\{wc t\}=we_t. \end{align*} Hence \begin{align*} \operatorname{span}_{\mathbb C}\{we_t:w\in S_n\} = \operatorname{span}_{\mathbb C}\{e_u:u\text{ is a }\lambda\text{-tableau}\} = E^\lambda. \end{align*} [guided] The theorem phrases the answer as the span of all polytabloids $e_u$ attached to tableaux $u$ of shape $\lambda$, whereas the preceding step identifies the image of $\mathbb C[S_n]c_t$ as the span of the translates $we_t$. To compare these descriptions, let $u$ be a $\lambda$-tableau. Because both $u$ and $t$ fill the same Young diagram with the entries $\{1,\dots,n\}$, there is a permutation $w\in S_n$ sending each entry of $t$ to the entry occupying the same box of $u$. Thus $u=wt$. The column stabilizer of $u$ is the conjugate of the column stabilizer of $t$: \begin{align*} C_u=wC_tw^{-1}. \end{align*} Indeed, a permutation preserves the columns of $u=wt$ setwise exactly when, after conjugating by $w^{-1}$, it preserves the columns of $t$ setwise. Since the sign character satisfies $\operatorname{sgn}(wcw^{-1})=\operatorname{sgn}(c)$ for every $c\in C_t$, the definition of the polytabloid gives \begin{align*} e_u=\sum_{d\in C_u}\operatorname{sgn}(d)\{du\}. \end{align*} Substitute $d=wcw^{-1}$ and $u=wt$. Then \begin{align*} e_u=\sum_{c\in C_t}\operatorname{sgn}(c)\{wcw^{-1}wt\}. \end{align*} Since $w^{-1}w$ is the identity permutation, \begin{align*} e_u=\sum_{c\in C_t}\operatorname{sgn}(c)\{wct\}=w\sum_{c\in C_t}\operatorname{sgn}(c)\{ct\}=we_t. \end{align*} Thus every polytabloid of shape $\lambda$ is a translate of $e_t$, and every translate of $e_t$ is a polytabloid of shape $\lambda$. Consequently the two spans are equal: \begin{align*} \operatorname{span}_{\mathbb C}\{we_t:w\in S_n\} = \operatorname{span}_{\mathbb C}\{e_u:u\text{ is a }\lambda\text{-tableau}\}. \end{align*} [/guided] [/step] [step:Conclude that the polytabloid span is an $S_n$-submodule isomorphic to $S^\lambda$] The subspace $E^\lambda$ is stable under the action of $S_n$: if $\sigma\in S_n$ and $e_u$ is a polytabloid, then by the computation in the previous step, \begin{align*} \sigma e_u=e_{\sigma u}\in E^\lambda. \end{align*} Hence $E^\lambda$ is an $S_n$-submodule of $M^\lambda$. Finally, restrict the $S_n$-module isomorphism $\Phi:\mathbb C[S_n]a_t\to M^\lambda$ to the submodule $\mathbb C[S_n]c_t$. Since $\Phi$ is injective on all of $\mathbb C[S_n]a_t$, its restriction \begin{align*} \Phi|_{\mathbb C[S_n]c_t}:\mathbb C[S_n]c_t\to E^\lambda \end{align*} is injective. The image of this restriction is $E^\lambda$ by the preceding two steps, so it is an isomorphism of $\mathbb C[S_n]$-modules. Since $S^\lambda=\mathbb C[S_n]c_t$ by definition, this gives \begin{align*} S^\lambda\cong E^\lambda. \end{align*} Thus the polytabloid span inside $M^\lambda$ realizes the Specht module. [/step]