Nilpotency class of the unitriangular matrix group

Question

I'm trying to compute the nilpotency class $c$ of unitriangular matrix group http://groupprops.subwiki.org/wiki/Unitriangular_matrix_group using the definition that $ [ ...[x_1, x_2 ] , x_3 ] ,..., x_{c + 1} ] = e $ for all $x_i \in G$.

I know the result is $ c =n -1 $ (from that page), and I think each step I should get a matrix with one more "layer" above the diagonal equal to $0$. But the computation seems very complicated. Is there a nice way to show this?

Any help is appreciated. Many thanks in advance.

Answer 1

You have to consider a formula in your link:

$$[e_{ij}(\lambda), e_{jk}(\mu)]=e_{ik}(\lambda\mu).$$ For simplicity, you may proceed with $n=3$, then $n=4$, and then in general.

Consider $n=4$, and $G=U(4,F)$. Then the generators of this group are $$e_{12}(\alpha), e_{13}(\alpha), e_{14}(\alpha), e_{23}(\alpha), e_{24}(\alpha), e_{34}(\alpha), \,\,\,\, \alpha\in F.$$

For nilpotency, it is sufficient to work on generators, and their commutators (in rough sense). By above formula, you can see that the commutators of the above generators are inside the normal subgroup $$ \begin{Bmatrix} \begin{bmatrix} 1 & 0 & * & *\\ & 1 & 0 & *\\ & & 1 & 0 \\ & & & 1 \end{bmatrix}\colon * \in F. \end{Bmatrix}. $$ Conversely, every element of this subgroup is a member of commutator group, since consider its generators $$ (*) \,\,\,\,\,\, e_{13}(\alpha), e_{14}(\alpha), e_{24}(\alpha).$$ They can be expressed as commutators: $$e_{13}(\alpha)=[e_{12}(\alpha), e_{23}(1)].$$ Try to express $e_{14}(\alpha), e_{24}(\alpha)$ as commutators.

Thus, above subgroup is equal to the commutator subgroup $[G,G]$.

To talk about commutators of length $3$, i.e. of the form $[[x,y],z]$, again, consider $[x,y]$ from generators of $[G,G]$ (in (*)) and the generators of $G$ as taken initially. You will notice, from commutator formula that the "three-commutators" (i.e. $[[x,y],z]$) will generate the normal subgroup

$$ \begin{Bmatrix} \begin{bmatrix} 1 & 0 & 0 & *\\ & 1 & 0 & 0\\ & & 1 & 0 \\ & & & 1 \end{bmatrix}\colon * \in F. \end{Bmatrix}. $$ This is sufficient to solve your problem.

Answer 2

An equivalent definition for a nilpotent group is that there are subgroups such that:

$$1=K_0\leq K_1\leq ...\leq K_{d-1}=G$$ $$[K_{i+1},G]\subset K_i$$

Let $G$ be the set of $d\times d$ unitriangular matrices.

We will build such a sequence of subgroups starting with the trivial group which is the identity matrix and adding "upper diagonal layers" to the matrices. Let $A_m\in K_m$. By this we mean:

$$A_{m}=I+N_m \quad \text{such that} \quad (N_m)_{ij}=0 \quad \text{if}\quad j\leq d-m+(i-1)$$

Before proving our main claim we need to verify a couple of things:

Claim: $N_m N_k=N_{m+k-d}$

Proof: $$(N_mN_k)_{ij}=\sum_l (N_m)_{il}(N_k)_{lj}=\sum_{l> d-m+(i-1)\quad j> d-k+(l-1)} (N_m)_{il}(N_{k})_{lj}$$ $$ \sum_{j-d+k+1 > l >d-m+(i-1) } (N_m)_{il}(N_k)_{lj}= \sum_{j-d+k \geq l \geq d-m+i } (N_m)_{il}(N_k)_{lj}$$ If $j-d+k< d-m+i$, there will be no elements to add and this entry will be zero. In other words, whenever $j \leq 2d-m-k+i-1=d-(m+k-d)+(i-1)$

We also note $n+m-d\leq (d-1)+m-d=m-1$, so multiplying by a given $N_n$ decreases the indice by at least one unity! Now we are ready to prove $K_m$ is a group.

Claim:$K_m$ is closed under matrix multiplication and matrix inverses:

Proof:

$$(I+N_m)(I+\tilde{N}_m)=I+N_m+\tilde{N}_m+N_{2m-d}$$ Clearly $2m-d\leq m+d-1-d=m-1$, so the multiplication remains in $K_m$. Futhermore the inverse can be computed directly, because $N_m^k=0$ for $k$ sufficiently large. (If we take $k=m$ it works). Then:

$$(I+N_m)^{-1}=I+\sum_{l=1}^{k-1}(-N_m)^l$$

Multiplying matrices in $N_m$ make us closer to the zero matrix so these powers of $N_m$ are no worry.

Next we prove the commutator identity we need:

Claim: $[K_{i+1},G]\subset K_i$

Proof: Let $k_{i+1}\in K_{i+1}$ and $g\in K_{d-1}=G$. This means we have $k_{i+1}=I+N_{i+1}$ and $g=I+\tilde{N}_{d-1}$

$$[k_{i+1},g]=(gk_{i+1})^{-1}(k_{i+1}g)=$$ $$(I+\tilde{N}_{d-1}+N_{i+1}+\tilde{N}_{d-1}N_{i+1})^{-1}(I+\tilde{N}_{d-1}+N_{i+1}+N_{i+1}\tilde{N}_{d-1} )$$

To keep our answer compendious let $u=\tilde{N}_{d-1}+N_{i+1}+\tilde{N}_{d-1}N_{i+1}$ and $v=\tilde{N}_{d-1}+N_{i+1}+N_{i+1}\tilde{N}_{d-1} $. This yields:

$$[k_{i+1},g]=(I+u)^{-1}(I+v)=\left(I+\sum_{l=1}^{k-1}(-u)^l\right)(I+v)=$$ $$I+\sum_{l=1}^{k-1}(-u)^l+v+\sum_{l=1}^{k-1}(-u)^l v=$$ $$ I+ (I+\sum_{l=1}^{k-2}(-u)^l)(v-u)-vu^{k-1}$$

We are done, because $v-u=N_{i+1}\tilde{N}_{d-1}-\tilde{N}_{d-1}N_{i+1}=N_{d+i-d}=N_{i}$ and $v u^{k-1}$ will decrease the indice of $u^{k-1}$ by one (so would multiplying by $u$), therefore $vu^{k-1}=0$