Baer correspondence between NT(3,p) and UT(3,p)

Throughout this article, $$p$$ is an odd prime number.

This article is about a particular instance of a Baer correspondence, and hence also a Lazard correspondence, between a Lie ring and a group, both of order $$p^3$$ and class two.

Note that most aspects of this article generalize to the following situations:


 * We replace $$\mathbb{F}_p$$ by a finite field $$\mathbb{F}_q$$ or an infinite field of characteristic not equal to 2.
 * Even more generally, we replace $$\mathbb{F}_p$$ by a ring whose characteristic is not divisible by 2. For instance, we may replace it by $$\mathbb{Z}/9\mathbb{Z}$$.

The parts that fail to generalize are the precise descriptions of subgroups and endomorphisms: in the more general cases, there are additional subgroups and endomorphisms.

The Lie ring
The Lie ring $$NT(3,p)$$ is niltriangular matrix Lie ring:NT(3,p). More explicitly, it is the Lie ring whose elements are $$3 \times 3$$ matrices over the prime field $$\mathbb{F}_p$$, with 0s on and below the diagonal, i.e., matrices of the form:

$$M(a,b,c) = \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \\\end{pmatrix}, \qquad a,b,c \in \mathbb{F}_p$$

The addition is defined as matrix addition and the Lie bracket is defined as $$[X,Y] = XY - YX$$ where the product is matrix multiplication. Explicitly:

The group
The group $$UT(3,p)$$ is unitriangular matrix group:UT(3,p). Explicitly, it is the group whose elements are $$3 \times 3$$ matrices over the prime field $$\mathbb{F}_p$$ with $$1$$s on the diagonal, $$0$$s below the diagonal, and arbitrary entries above the diagonal, i.e., matrices of the form:

$$N(a,b,c) = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \\\end{pmatrix}, \qquad a,b,c \in \mathbb{F}_p$$

The multiplication is matrix multiplication and the identity element and inverse are the usual identity element and inverse. Explicitly:

From Lie ring to group
The bijection given by the Baer correspondence is the exponential map for matrices, explicitly:

$$X \mapsto I + X + \frac{X^2}{2}$$

where the addition and multiplication are carried out as matrices. Note that higher powers of $$X$$ do not appear because $$X^3$$ becomes zero.

In terms of our notation, this is the map:

$$\begin{pmatrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \\\end{pmatrix} \mapsto \begin{pmatrix} 1 & a & b + (ac)/2 \\ 0 & 1 & c \\ 0 & 0 & 1 \\\end{pmatrix}$$

Or, in our shorthand:

$$\! M(a,b,c) \mapsto N(a,b+(ac)/2,c)$$

From group to Lie ring
The bijection given by the Baer correspondence is the logarithm map for matrices, explicitly:

$$X \mapsto (X - I) - \frac{(X - I)^2}{2}$$

where the addition and multiplication are carried out as matrices. Note that higher powers of $$(X - I)$$ do not appear because $$(X- I)^3$$ becomes zero.

In terms of our notation, this is the map:

$$\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \\\end{pmatrix} \mapsto \begin{pmatrix} 0 & a & b - (ac)/2) \\ 0 & 0 & c \\ 0 & 0 & 0 \\\end{pmatrix}$$

In our shorthand:

$$\! N(a,b,c) \mapsto M(a,b - (ac)/2,c)$$

Central series
Both the Lie ring and the group are nilpotent of class equal to 2. In fact, they both have the lower central series and upper central series coincide, i.e., the center and derived subgroup are the same for the group, and the center and derived subring are the same for the Lie ring. Specifically, we have:

$$Z(NT(3,p)) = [NT(3,p),NT(3,p)] = \{ M(0,b,0) : b \in \mathbb{F}_p \} $$ is an abelian Lie ring whose additive group is a group of prime order. $$Z(UT(3,p)) = [UT(3,p),UT(3,p)] = \{ N(0,b,0) : b \in \mathbb{F}_p \}$$ is a group of prime order.

Further, under the bijection, $$M(0,b,0) \leftrightarrow N(0,b,0)$$, so the bijection restricts to an isomorphism on the center with the obvious identification.

Correspondence between subgroups and subrings
The table below lists the five different automorphism classes of subgroups and the corresponding automorphism classes of subrings, with comments on how the bijection works for each of these.