I am currently working on Bruce Sagan's The Symmetric Group.
In the proof of Corollary 2.4.2, the book says
Let $t = t^\lambda$ be a $\lambda$-tableau and $s = s^{\mu}$ be a $\mu$-tableau, where $\lambda, \mu$ are partitions of an integer $n$. If $\lambda = \mu$, then we must have $\{s\} = \pi\{t\}$ for some $\pi \in C_t$ by the same argument that established the dominance lemma.
The dominance lemma was proven in the following way:
Let $t^{\lambda}$ and $s^{\mu}$ be tableaux of shape $\lambda$ and $\mu$, respectively. If, for each index $i$, the elements of row $i$ of $s^{\mu}$ are all in different columns in $t^{\lambda}$, then $\lambda \trianglerighteq \mu$.
Proof: By hypothesis, we can sort the entries in each column of $t^{\lambda}$ so that the elements of rows $1,2,...i$ of $s^{\mu}$ all occur in the first $i$ rows of $t^{\lambda}$. Thus $$\lambda_1 + \lambda_2 + ... \lambda_i = \text{number of elements in the first $i$ rows of $t^{\lambda}$} \geq \text{number of elements of $s^{\mu}$ in the first $i$ rows of $t^{\lambda}$ } = \mu_1 + \mu_2 + .... + \mu_i.$$
However, I cannot see, why this argument should prove that we can find such a $\pi \in C_t$. Take for example
$\text{1 2 3}$
$\text{4 5}$
$\text{6}$
and
$\text{4 5 6}$
$\text{1 2}$
$\text{3}$
for $t$ and $s$ respectively. When we work with $tabloids$, we can interchange the elements of the rows as we like. So if the required $\pi$ exists, then clearly $6$ must be sent to $3$, as they are the only elements in row 3 of the respective tableaux. But then, they must be in the same column, giving us tableaux
$\text{3 1 2}$
$\text{4 5}$
$\text{6}$
and
$\text{6 4 5}$
$\text{1 2}$
$\text{3}$.
But, as we also need the $1$s and the $4$s need to be in the same column, we are forced to write
$\text{3 1 2}$
$\text{5 4}$
$\text{6}$
and
$\text{6 4 5}$
$\text{2 1}$
$\text{3}$,
leaving us no chance to exchange $2$ and $5$.
Am I making a mistake or is there no $\pi \in C_t$? If there was one, how would I go in proving it?
Thank you very much for your help!