hism as a linear combination of semistandard homomorphisms.

We now give another result which will help us in showing that a linear combination of row-standard homomorphisms is non-zero without having to go through the full process of expressing it as a linear combination of semistandard homomorphisms. This concerns the *dominance order* on tableaux.

Suppose $\mu$ is a partition, and $S,T$ are row-standard $\mu$-tableaux of the same type. We say that $S$ dominates $T$, and write $S\dom T$, if it is possible to get from $S$ to $T$ by repeatedly swapping an entry of $S$ with a larger entry in a lower row (and re-ordering with each row). We warn the reader that this is not quite the same as the dominance order described in [@j2 13.8]. For example, the dominance order on ${\calt_{\hspace{-2pt}\operatorname{r}}}\left((3,2),(2^2,1)\right)$ is given by the following Hasse diagram.

$$\begin{tikzpicture}
\draw(0.5,0.5)--++(0,2.5)--++(2,2)--++(-2,2)--++(-2,-2)--++(2,-2);
\tyoung(0cm,0.5cm,223,11)
\tyoung(0cm,3cm,123,12)
\tyoung(2cm,5cm,122,13)
\tyoung(-2cm,5cm,113,22)
\tyoung(0cm,7cm,112,23)
\end{tikzpicture}$$

Now we have the following lemma.

\[semidom\] Suppose $\mu$ is a partition of $n$ and $\la$ a composition of $n$, and $T\in{\calt_{\hspace{-2pt}\operatorname{r}}}(\mu,\la)$. If we write $${\hat\Theta_{T}}=\sum_{S\in{\calt_{\hspace{-2pt}0}}(\mu,\la)}a_S{\hat\Theta_{S}},$$ then $a_S\neq0$ only if $S\dom T$.

For this proof, we adopt the set-up of [@j2 §13]. If we fix a $\mu$-tableau $t$ (of type $(1^n)$), then $t$ determines a natural bijection between the set of $\la$-tabloids and the set of $\mu$-tableaux of type $\la$. We identify tabloids with tableaux according to this bijection. We also let $e_t$ denote the polytabloid indexed by $t$, which generates $S^\mu$. Given $\la$-tableaux $U,V$, we write $U{\stackrel{\text{row}}\longleftrightarrow}V$ if $V$ can be obtained from $U$ by permuting entries within rows, and we define $U{\stackrel{\text{column}}\longleftrightarrow}V$ similarly. Now given $U\in{\calt_{\hspa