Algebra group structures for cyclic group:Z4

There exists a unique way of making cyclic group:Z4 into an algebra group over field:F2. It is not an algebra group over any other field. This is unique in a strict sense, not merely up to isomorphisms.

Multiplication table (structure constants)
The algebra is two-dimensional. It has the following multiplication table, with basis elements $$a,b$$. Note that Adjoint group of a radical ring is abelian iff the radical ring is commutative‎, so the algebra group is commutative, and we don't have to worry about the order of multiplication:

Verification of properties

 * $$N$$ is associative: All products of length three or more involving basis elements are zero, hence, by linearity, all products of length three or more are zero.
 * $$N$$ is nilpotent: All products of length three or more involving basis elements are zero, hence, by linearity, all products of length three or more are zero.
 * The algebra group for $$N$$ is cyclic of order 4: Indeed, $$(1 + a)^2 = 1 + b$$ and $$(1 + b)^2 = 1$$, so $$1 + a$$ is an element of order four (see also powering map by field characteristic is same in algebra and algebra group).

Description as subalgebra of niltriangular matrix Lie algebra
The algebra can be realized explicitly as a subalgebra of niltriangular matrix Lie algebra:NT(3,2) as follows:

$$a = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\\end{pmatrix}, \qquad b = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\\end{pmatrix}$$