Nilpotency of higher dimension Heisenberg groups I'm trying to understand higher dimension Heisenberg groups, defined as:
$$H^{2k+1} = \langle a_1,\ldots,a_k,b_1,\ldots,b_k,c \mid [a_i,b_i] = c, [a_i,c] = [b_i,c]=1, [a_i,a_j] =[b_i,b_j] = 1, i \neq j\rangle$$
Why are they nilpotent? What are their nilpotency classes? I'm looking for an answer from the group-theoretic point of view.
Edited to correct the relations.
Thanks.
 A: First we compute $[G,G]$. Since $[a_1,b_1] = c$, we get that $\langle c \rangle \leq [G,G]$. Since $c$ commutes with $a_i$ and $b_i$ (and $c$), $\langle c \rangle$ is a normal subgroup. In $G/\langle c \rangle$ we get that $[\bar a_i, \bar b_i ] = \bar c = \bar 1$ and the other commutators were all $1$ already, so we get that all the generators commute, so the group $G/\langle c\rangle$ is abelian. By definition of $[G,G]$, we have that $[G,G] \leq \langle c\rangle$ since $G/\langle c \rangle$ is an abelian quotient.
Hence $[G,G] = \langle c \rangle$.
Now we compute the elements of $G$: Every element of $G$ has an expression as $a_1^{e_1} a_2 ^{e_2} \cdots a_k^{e_k} b_1^{f_1} b_2^{f_2} \cdots b_k^{f_k} c^g$ for integers $e_i, f_i, g$. This follows from the relations that say $b_j a_i = a_i b_j$ or $b_i a_i = a_i b_i c^{\pm1}$ so we can alphabetize the $a$ versus $b$ if we are willing to create new copies of $c$. Of course $c$ commutes with everything so we can move the $c$ to the right, and the $a$ commute amongst themselves so we can order them numerically, and similarly the $b$.
Now we compute $Z(G)$ by checking EVERY element. $c^g$ works clearly, so $\langle c \rangle \leq Z(G)$. Now suppose $z \in Z(G)$ has been written as above with the $e_i, f_i,g$. If $e_i \neq 0$ for some $i$, then consider $b_i z$ versus $z b_i$. Since $b_i$ commutes with everything in $z$ except $a_i$ we get that $z'=z b_i$ has all the same powers, except $f_i' = f_i + 1$. On the other hand, $z'' = b_i z$ has two changes: $f_i'' = f_i+1$ and $g''=g\pm1$ based on $b_i a_i = a_i b_i c^{\pm1}$. Hence such a $z$ is not really in the center. Hence if $z$ really is in the center than $e_i=0$ for all $i$. A similar argument shows $f_i=0$ as well. Hence $z \in \langle c \rangle$ and $Z(G) = \langle c \rangle$.
Now $G/Z(G)$ is abelian, so $Z^2(G)/Z(G) = G/Z(G)$ and $Z^2(G) = G$ so $G$ has nilpotency class 2. Similarly, $[G,G]$ is central, so $[G,G,G]=1$ and $G$ has nilpotency class 2.
Note that all we needed was that $[G,G] \leq \langle c \rangle \leq Z(G)$ to calculate the nilpotency class, I included the exact calculations because you asked.
