[proofplan] We prove the [dominance order](/page/Dominance%20Order) condition by starting with an arbitrary [semistandard Young tableau](/page/Semistandard%20Young%20Tableau) $T$ of shape $\lambda$ and [content](/page/Content%20of%20a%20Tableau) $\mu$. For each $k$, we count the boxes whose entries are at most $k$ in two ways: by content, their number is $\mu_1+\cdots+\mu_k$, while column-strictness forces all such boxes to lie in the first $k$ rows, whose total size is $\lambda_1+\cdots+\lambda_k$. When $\mu=\lambda$, the same counting argument gives equality for every $k$, forcing the boxes in row $i$ to have entry exactly $i$; this produces exactly one semistandard tableau. [/proofplan] [step:Use column strictness to locate all entries at most $k$ in the first $k$ rows] Assume $K_{\lambda\mu} \neq 0$. Choose a [semistandard Young tableau](/page/Semistandard%20Young%20Tableau) $T$ of shape $\lambda$ and [content](/page/Content%20of%20a%20Tableau) $\mu$. For each integer $k \geq 1$, define \begin{align*} A_k := \{b \in \lambda : T(b) \leq k\}, \end{align*} where $b \in \lambda$ denotes a box in the Young diagram of $\lambda$. We claim that every box in $A_k$ lies in one of the first $k$ rows. Indeed, let $b$ be a box in row $r$. Since $T$ is column-strict, the entries in the column containing $b$ strictly increase from top to bottom. Therefore the entry in row $r$ of that column is at least $r$, because there are $r$ positive integers in that column up to and including row $r$, and they are strictly increasing. Hence if $T(b) \leq k$, then $r \leq k$. [guided] Fix an integer $k \geq 1$. We want to compare the number of entries of $T$ that are at most $k$ with the number of boxes available in the first $k$ rows of the shape $\lambda$. Define \begin{align*} A_k := \{b \in \lambda : T(b) \leq k\}. \end{align*} This is the set of boxes whose entries belong to $\{1,\dots,k\}$. The key point is that column-strictness prevents such a box from appearing below row $k$. Let $b$ be a box in row $r$. In the column containing $b$, the entries strictly increase as one moves downward. Since all entries of a semistandard Young tableau are positive integers, the entry in the first row of that column is at least $1$, the entry in the second row is at least $2$, and continuing inductively, the entry in row $r$ is at least $r$. Therefore \begin{align*} T(b) \geq r. \end{align*} If $b \in A_k$, then $T(b) \leq k$, so \begin{align*} r \leq T(b) \leq k. \end{align*} Thus every box in $A_k$ lies in one of the first $k$ rows. [/guided] [/step] [step:Count the entries at most $k$ to obtain the dominance inequalities] By the content condition, the number of boxes in $A_k$ is \begin{align*} |A_k| = \sum_{i=1}^{k} \mu_i. \end{align*} By the previous step, $A_k$ is contained in the union of the first $k$ rows of $\lambda$, whose number of boxes is \begin{align*} \sum_{i=1}^{k} \lambda_i. \end{align*} Therefore, for every $k \geq 1$, \begin{align*} \sum_{i=1}^{k} \mu_i = |A_k| \leq \sum_{i=1}^{k} \lambda_i. \end{align*} By the definition of the [dominance order](/page/Dominance%20Order), this is exactly $\lambda \unrhd \mu$. [/step] [step:Show that content equal to the shape forces one tableau] Now assume $\mu=\lambda$. Let $T$ be a semistandard Young tableau of shape $\lambda$ and content $\lambda$. For each $k \geq 1$, define \begin{align*} A_k := \{b \in \lambda : T(b) \leq k\}. \end{align*} As above, $A_k$ is contained in the first $k$ rows. Since the content is $\lambda$, \begin{align*} |A_k| = \sum_{i=1}^{k} \lambda_i, \end{align*} which is exactly the number of boxes in the first $k$ rows. Hence $A_k$ is the set of all boxes in the first $k$ rows. It follows that the boxes in row $i$ lie in $A_i$ but not in $A_{i-1}$. Therefore every box in row $i$ has entry exactly $i$. Thus any such tableau must be the filling with row $i$ filled entirely by $i$. [/step] [step:Verify that the forced filling is semistandard and conclude uniqueness] Define $T_\lambda$ to be the filling of the Young diagram of $\lambda$ given by \begin{align*} T_\lambda(b) := i \end{align*} for every box $b$ in row $i$. Each row is weakly increasing because it is constant. Each column is strictly increasing because the entry in row $i+1$ is $i+1$, which is greater than the entry $i$ directly above it. The content of $T_\lambda$ is $\lambda$, since row $i$ contains exactly $\lambda_i$ boxes, all labelled $i$. The previous step shows that every semistandard Young tableau of shape $\lambda$ and content $\lambda$ must equal $T_\lambda$, while this step shows that $T_\lambda$ is semistandard. Hence there is exactly one such tableau: \begin{align*} K_{\lambda\lambda} = 1. \end{align*} [/step]