Algebra group

General definition
Let $$N$$ be an associative algebra over a field $$F$$ with the property that for any $$x \in N$$, there exists $$y \in N$$ such that $$x + y + xy = 0$$. (in other words, $$N$$ is a radical ring).

Then, the algebra group corresponding to $$N$$ is the adjoint group of $$N$$, i.e., the group whose elements are formally:

$$\{ 1 + x | x \in N\}$$

where the multiplication is defined as:

$$(1 + x)(1 + y) = 1 + (x + y + xy)$$

There are two interpretations of this:


 * 1) We can think of the algebra group as living inside the unitization $$N + F$$ as the coset of $$N$$ for $$1 \in F$$.
 * 2) Alternatively, since there is a bijection between $$N$$ and its algebra group, we can think of the algebra group as simply the set $$N$$ with group operation $$x * y := x + y +  xy$$. This is based on the idea of the multiplicative formal group law. However, unlike other formal group laws, the multiplicative formal group law is valid even in non-commutative settings, and that is necessary for this definition.

Nilpotent case
If $$N$$ is an associative algebra over a field $$F$$ with the property that every element of $$N$$ is nilpotent, then it is true that for any $$x \in N$$, there exists $$y \in N$$ such that $$x + y + xy = 0$$. Thus, we can always define the algebra group corresponding to any such algebra $$N$$.

Finite field case
If $$F$$ is a finite field and $$N$$ is a finite-dimensional associative algebra over $$F$$, the following are equivalent:


 * 1) For any $$x \in N$$, there exists $$y \in N$$ such that $$x + y + xy = 0$$.
 * 2) Every element of $$N$$ is nilpotent.
 * 3) $$N$$ is a nilpotent algebra, i.e., there is some fixed length such that all products of that length of elements in $$N$$ equal zero.

Thus, for a finite-dimensional associative algebra over a finite field $$F$$, we talk of its algebra group only if it satisfies these equivalent conditions.

In the context of an abstract group
A group of prime power order which can be realized as an algebra group any finite field is termed an algebra group. Note that an algebra group over $$\mathbb{F}_q$$, where $$q$$ is a power of a prime $$p$$, must also be an algebra group over the prime field $$\mathbb{F}_p$$. Thus, saying that a finite $$p$$-group is an algebra group is equivalent to saying that it is an algebra group over $$\mathbb{F}_p$$.

Uniqueness and non-uniqueness of the algebra
Note that the nilpotent associative algebra up to isomorphism determines the algebra group up to isomorphism. However, it is not a priori clear whether the group up to isomorphism determines the nilpotent associative algebra up to isomorphism. It turns out that it does not, but there are important restrictions on the structure of the algebra that can be deduced from the group structure.


 * Non-isomorphic nilpotent associative algebras may have isomorphic algebra groups
 * Powering map by field characteristic is same in algebra and algebra group
 * Lazard Lie ring of adjoint group of a radical ring equals associated Lie ring of the radical ring under suitable nilpotency assumptions

Necessary conditions for being an algebra group
Given a finite p-group, it is not necessarily an algebra group over $$\mathbb{F}_p$$. Below are some things we can deduce if it is an algebra group:


 * Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size: Note that this gives a complete characterization of algebra groups, at least in the finite case. What it says is that any algebra group over $$\mathbb{F}_q$$ of order $$q^m$$ is isomorphic to an algebra subgroup of the unitriangular matrix group $$UT(m+1,q)$$.
 * Algebra group implies power degree group for field size