Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / A / 1

Define $L_1=L$ and $L_{n+1}=[L,L_n]$, where brackets of subspaces mean their linear span. Each term is an ideal. Inductively, if $L_n$ is an ideal, the Jacobi identity givesso $L_{n+1}$ is an ideal as well. Also $L_{n+1}\subseteq L_n$. Nilpotence means that this descending series eventually reaches zero; class $c$ means $L_{c+1}=0$ with $L_c\ne 0$ for nonzero $L$.

The idealizer here is the normalizer ${\mathrm{Id}}_L(A)=\{x\in L:[x,A]\subseteq A\}$. Since $A$ is a subalgebra it contains its own brackets, so $A\subseteq {\mathrm{Id}}_L(A)$.

Because $L_1⊈A$ and the lower central series terminates, choose $j$ with $L_j⊈A$ but $L_{j+1}\subseteq A$. Take $x\in L_j\setminus A$. Then $[x,A]\subseteq [L_j,L]=L_{j+1}\subseteq A$. Thus $x$ belongs to the idealizer and not to $A$, proving $\boxed{A<{\mathrm{Id}}_L(A)}$.

Start with $I_0=I$ and set $I_{k+1}=[L,I_k]$. Every term lies in $I$, and $I_k\subseteq L_{k+1}$, so the sequence eventually vanishes. Choose its last nonzero term $I_j$. Then $[L,I_j]=0$, hence $0\ne I_j\subseteq I\cap Z(L)$. Therefore $\boxed{I\cap Z(L)\ne \{0\}}$.

Induct on $i$, proving the claim for all $j$ simultaneously. The case $i=1$ is the definition $[L,L_j]=L_{j+1}$. If it holds for $i$, Jacobi givesBoth uses of the induction hypothesis have first index $i$, so the induction is valid. Thus $\boxed{[L_i,L_j]\subseteq L_{i+j}}$ for all positive indices.

Let $M=L_r$ and write $M_k$ for its own lower central series. Part (c) proves inductively that $M_k\subseteq L_{rk}$: it holds at $k=1$, and $[M,M_k]\subseteq [L_r,L_{rk}]\subseteq L_{r(k+1)}$. If $L$ has class less than $rs$, then $L_{rs}=0$, so $M_s=0$. Consequently $\boxed{\mathrm{class}(L_r)<s}$, with the zero algebra understood to have class zero.