Conjugation reverses dominance

Fix $k\ge 1$. Double counting the nodes of $[\lambda ]$ in its first $k$ columns gives ${\sum }_{j\le k}{\lambda }_j^{\mathrm{\prime }}={\sum }_i\min ({\lambda }_i,k)=n-{\sum }_i\max (0,{\lambda }_i-k)$; and because $\lambda$ is weakly decreasing, ${\sum }_i\max (0,{\lambda }_i-k)={\max }_{r\ge 0}({\sum }_{i\le r}{\lambda }_i-rk)$, this maximum being attained at the finite index $r={\lambda }_{k+1}^{\mathrm{\prime }}$ (with $r=0$ when $k\ge {\lambda }_1$, using zero-padding). Hence ${\sum }_{j\le k}{\lambda }_j^{\mathrm{\prime }}=n-{\max }_{r\ge 0}({\sum }_{i\le r}{\lambda }_i-rk)$ holds for every $k\ge 1$, and both sides vanish when $\lambda =\mathrm{\varnothing }$.

Assume $\lambda ⊵\mu$, and fix $k\ge 1$. By [L1] one has ${\sum }_{i\le r}{\lambda }_i\ge {\sum }_{i\le r}{\mu }_i$ for every $r\ge 0$, the case $r=0$ reading $0\ge 0$, so the maximum appearing in step 1.1 for $\lambda$ is at least the corresponding maximum for $\mu$; subtracting both from $n$ gives ${\sum }_{j\le k}{\lambda }_j^{\mathrm{\prime }}\le {\sum }_{j\le k}{\mu }_j^{\mathrm{\prime }}$. As $k\ge 1$ was arbitrary, ${\mu }^{\mathrm{\prime }}⊵{\lambda }^{\mathrm{\prime }}$.

Conversely assume ${\mu }^{\mathrm{\prime }}⊵{\lambda }^{\mathrm{\prime }}$. The partitions ${\lambda }^{\mathrm{\prime }}$ and ${\mu }^{\mathrm{\prime }}$ of $n$ are a pair of partitions of the same integer, so the implication of step 2.1 applies to them and yields $({\lambda }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}⊵({\mu }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$; by the involution $({\lambda }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\lambda$ and $({\mu }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\mu$ of [L3] this is $\lambda ⊵\mu$.

Step 2.1 proves the forward implication and step 3.1 the reverse one, so for all partitions $\lambda ,\mu ⊢n$ one has $\lambda ⊵\mu$ if and only if ${\mu }^{\mathrm{\prime }}⊵{\lambda }^{\mathrm{\prime }}$. ∎