# Nonabelian infinite nilpotent groups

Can you give examples of nonabelian infinite nilpotent groups?

Here's what I got so far:

1. The Heisenberg group.
2. The free nilpotent group of class $s$ (thanks Arturo for your comment here).
3. The group of (some) symmetries of polynomials of degree up to by $s$ generated by the symmetry which adds a constant polynomial (for each constant polynomial) and translation of the argument of the polynomial by a scalar (for each scalar).

(got the last 2 examples from http://terrytao.wordpress.com/2009/12/21/the-free-nilpotent-group/).

I'm looking for more examples of such groups. Infinite nonabelian groups which are not nilpotent, but have a nilpotent subgroup of finite index are also of interest to me.

• There is no such thing as a "free nilpotent group"; there are "free nilpotent group of class $c$", though. Tao is describing these groups. – Arturo Magidin Mar 2 '12 at 21:33
• I'm an undergrad student in CS trying to think independently of some problems, and these objects may be able to help me. Even if not, it would be nice to see the examples. – Jackie Mar 2 '12 at 21:33
• @Arturo: Thanks. Question changed accordingly. – Jackie Mar 2 '12 at 21:37
• upper triangular matrices with $1$'s on the diagonal – user8268 Mar 2 '12 at 23:00

1. Let $R$ be a ring (not necessarily commutative, not necessarily with identity); if $n\gt 0$, then we let $R^{(n)}$ be the set of all sums of products of $n$ elements of $R$; then $R^{(n)}$ is a subring of $R$, and $R^{(n)}=0$ if and only if every product of $n$ elements of $R$ is equal to $0$. If $R^{(n)}=0$ for some $n\gt 0$, we say that $R$ is a nilpotent ring.
Now let $S$ be a ring with identity, and assume that $R$ is a subring of $S$ that is a nilpotent ring (in particular, $R$ does not contain the identity). Let $U$ be the set of all elements of the form $1+x$ with $x\in R$. Then $U$ is a group under the operation given by multiplication in $S$: that is, given $1+x$ and $1+y$ in $U$, then $$(1+x)(1+y) = 1 + (x + y + xy) \in U$$ and $$(1+x)^{-1} = 1 + (-x + x^2 - x^3 + x^4 -\cdots + (-1)^{n-1}x^{n-1})\in U,$$ where $x^n = 0$.
Then $U$ is a nilpotent group, and if we let $U_i = R^{(i)}$, then $$1 = U_n \leq U_{n-1}\leq\cdots\leq U_1 = U$$ is a central series for $U$.
For example, let $S$ be the ring of $n\times n$ matrices over an infinite commutative ring with identity $T$, and let $R$ be the set of upper zero triangular matrices; then $R$ is nilpotent, and the corresponding $U$ is the group of upper unitriangular matrices over $T$, which is nilpotent of class exactly $n-1$. (This generalizes the Heisenberg group).
2. Let $A$ be a nontrivial abelian group, let $D=A\times A$, and let $\delta\colon D\to D$ be given by $\delta(a,b) = (a,ab)$. Let $G=D\rtimes\langle \delta\rangle$. Then $G$ is nilpotent of class $2$. If $A$ is infinite, so is $G$.