Coinvariants detect vanishing of subquotients of bounded weight, and padded irreducibles stabilise

Apply [F1] with $\mu =\lambda [n]⊢n$ and $k_0=n-a$: the $S_a$-module $(V(\lambda {)}_n{)}_{S_{n-a}}$ is the sum of the irreducibles $V_{\nu }$, $\nu ⊢a$, over the partitions obtained from $\mu$ by removing one box from each of $k_0=n-a$ different columns, all multiplicities at most one.

The partition $\mu =({\mu }_1,{\mu }_2,\dots )$ has exactly ${\mu }_1$ columns, so no such removal exists when $n-a>{\mu }_1$; conversely, for every $0\le k_0\le {\mu }_1$, removing the bottom box of each of the $k_0$ rightmost columns is a valid horizontal-strip removal and leaves the partition $\nu$ with column heights ${\mu }_j^{\mathrm{\prime }}-1$ for those columns. Since ${\mu }_1=n-m$, the condition $n-a>{\mu }_1$ is $a<m=\mathrm{\mid }\lambda \mathrm{\mid }$, so $(V(\lambda {)}_n{)}_{S_{n-a}}=0$ if and only if $a<\mathrm{\mid }\lambda \mathrm{\mid }$, proving (i).

If $n-a={\mu }_1$, the strip has one box in each of the ${\mu }_1$ columns of $\mu$, and each removed box must be the bottom box of its column for the complement to remain a partition; the resulting partition has column heights ${\mu }_j^{\mathrm{\prime }}-1$, namely $\lambda$ (because ${\mu }_j^{\mathrm{\prime }}-1=\mathrm{\#}\{i:{\lambda }_i\ge j\}$ in the padded shape $\mu =(n-m,{\lambda }_1,{\lambda }_2,\dots )$). Hence $(V(\lambda {)}_n{)}_{S_{n-a}}\cong V_{\lambda }$, which is (ii).

For (iii), let $\nu ⊢a$ and compare the two constituent conditions. By step 1.1, $\nu$ occurs in $(V(\lambda {)}_n{)}_{S_{n-a}}$ if and only if $\nu$ is obtained from $\mu =(n-m,{\lambda }_1,{\lambda }_2,\dots )$ by removing one box from each of $n-a$ different columns, i.e. ${\mu }_i\ge {\nu }_i\ge {\mu }_{i+1}$ for all $i$; these inequalities read $n-m\ge {\nu }_1\ge {\lambda }_1$ and ${\lambda }_{i-1}\ge {\nu }_i\ge {\lambda }_i$ for $i\ge 2$. By [F1] applied to the induction from $S_m\times S_{a-m}$ with $m=\mathrm{\mid }\lambda \mathrm{\mid }$ (part (i) of that lemma), $\nu$ occurs in ${\mathrm{Ind}}_{S_m\times S_{a-m}}^{S_a}(V_{\lambda }⊠k)$ if and only if $\nu$ is obtained from $\lambda$ by adding one box in each of $a-m$ different columns, i.e. ${\nu }_i\ge {\lambda }_i\ge {\nu }_{i+1}$ for all $i$. The two systems differ only in the bound ${\nu }_1\le n-m$, which is automatic in the range $n\ge a+{\lambda }_1$ because ${\nu }_1\le {\lambda }_1+(a-m)\le n-m$; hence the two modules have exactly the same constituents, each with multiplicity at most one, and they are isomorphic.

For (iv), suppose first that $W_n\mathrm{\ne }0$. Choose $0\mathrm{\ne }w\in W_n$ and let $U\subseteq W_n$ be the $k$-span of the finite $S_n$-orbit of $w$; then $U$ is a nonzero finite-dimensional $k{S}_n$-submodule. By [F2] $U$ has a simple submodule, which by the classification is $V(\eta {)}_n=V_{\rho }$ for some partition $\rho =\eta [n]$ with $\eta ⊢r$. As $V_{\rho }$ is a constituent of $V_n$ and $\mathrm{weight}(V)\le a$, we have $r=\mathrm{\mid }\eta \mathrm{\mid }\le a$ (Weight of an FI-module over a field of characteristic zero). Put $k_0=n-a\le n-r$, and apply step 2.1 with ${\lambda }^{\mathrm{\prime }}=\eta$: since $a\ge r$, the coinvariants $(V(\eta {)}_n{)}_{S_{n-a}}$ are nonzero. The inclusion $V(\eta {)}_n\hookrightarrow U\hookrightarrow W_n$ induces an injection on coinvariants by exactness [F3], so $(W_n{)}_{S_{n-a}}\mathrm{\ne }0$. Conversely, if $W_n=0$ then $(W_n{)}_{S_{n-a}}=0$ trivially. This proves (iv) and completes the lemma.