Algebra group structures for direct product of Z4 and Z2

There exist (at least) three isomorphism classes of nilpotent associative algebras $$N_1,N_2,N_3$$ over field:F2 such that both the algebras have algebra group isomorphic to direct product of Z4 and Z2.

Of these, $$N_1$$ is a direct product of algebra groups, whereas $$N_2$$ and $$N_3$$ are not.

Multiplication table (structure constants)
We choose basis letters $$a,b,c$$ for $$N_1$$, and give it the following multiplication table. The row element is multiplied on the left and the column element on the right (though this is irrelevant since multiplication is commutative anyway).

Verification of properties

 * $$N_1$$ is associative: By the linearity, it suffices to check associativity on basis triples. It's easy to see from the multiplication table that all products for basis triples are zero, so associativity holds.
 * $$N_1$$ is nilpotent: All products of length three or more are zero, so the algebra is nilpotent.
 * The algebra group of $$N_1$$ is isomorphic to direct product of Z4 and Z2: We can verify that $$1 + a$$ squares to $$1 + b$$ and has order four. The element $$1 + c$$ commutes with both $$1 + a$$ and $$1 + b$$ and has order two, so we get an internal direct product of $$\langle 1 + a \rangle$$ and $$\langle 1 + c \rangle$$, which is isomorphic to direct product of Z4 and Z2.

Description as subalgebra of niltriangular matrix Lie algebra
The algebra can be realized explicitly as a subalgebra of niltriangular matrix Lie ring:NT(4,2) as follows:

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

Multiplication table (structure constants)
We choose basis letters $$x,y,z$$ for $$N_2$$, and give it the following multiplication table. The row element is multiplied on the left and the column element on the right (though this is irrelevant since multiplication is commutative anyway).

Verification of properties

 * $$N_2$$ is associative: By the linearity, it suffices to check associativity on basis triples. If a triple involves any occurrence of $$y$$ or $$z$$, the product either way is zero. For the triple $$xxx$$, the product either way is $$z$$.
 * $$N_2$$ is nilpotent: All products of length four or more are zero, so the algebra is nilpotent. This is because as noted above, all basis products of length three are either 0 or $$z$$, so all basis products of length four are zero.
 * The algebra group of $$N_2$$ is isomorphic to direct product of Z4 and Z2: We can verify that $$1 + x$$ squares to $$1 + y$$ and has order four. The element $$1 + z$$ commutes with both $$1 + x$$ and $$1 + y$$ and has order two, so we get an internal direct product of $$\langle 1 + x \rangle$$ and $$\langle 1 + z \rangle$$, which is isomorphic to direct product of Z4 and Z2.

Description as subalgebra of niltriangular matrix Lie algebra
The algebra can be realized explicitly as a subalgebra of niltriangular matrix Lie ring:NT(4,2) as follows:

$$x = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\\end{pmatrix}, \qquad y = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{pmatrix},\qquad z = \begin{pmatrix}  0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{pmatrix}$$

Multiplication table (structure constants)
We choose basis letters $$g,h,k$$ for $$N_3$$, and give it the following multiplication table. The row element is multiplied on the left and the column element on the right (though this is irrelevant since multiplication is commutative anyway).

Verification of properties

 * $$N_3$$ is associative: By the linearity, it suffices to check associativity on basis triples. In fact, all products on basis triples parenthesized either way are zero.
 * $$N_3$$ is nilpotent: All products of length three or more are zero, so the algebra is nilpotent. This is because as noted above, all basis products of length three are zero.
 * The algebra group of $$N_3$$ is isomorphic to direct product of Z4 and Z2: We can verify that $$1 + g$$ squares to $$1 + h$$ and has order four. The element $$1 + k$$ commutes with both $$1 + g$$ and $$1 + h$$ and has order two, so we get an internal direct product of $$\langle 1 + xg \rangle$$ and $$\langle 1 + k \rangle$$, which is isomorphic to direct product of Z4 and Z2.

Description as subalgebra of niltriangular matrix Lie algebra
The algebra can be realized explicitly as a subalgebra of niltriangular matrix Lie ring:NT(4,2) as follows:

$$g = \begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{pmatrix}, \qquad h = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{pmatrix},\qquad k = \begin{pmatrix}  0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\\end{pmatrix}$$

Non-isomorphism of these algebras
$$N_2$$ is clearly not isomorphic to either $$N_1$$ or $$N_3$$ because it has a nonzero product of length three, unlike the other two.

Distinguishing $$N_1$$ and $$N_3$$ is harder. Here's a way:


 * Both $$N_1$$ and $$N_3$$ have a unique element that occurs as a product. The element is $$b$$ in $$N_1$$ and $$h$$ in $$N_3$$.
 * In $$N_1$$, the product of any two elements, both of which square to $$b$$, equals $$b$$. However, in $$N_3$$, the product of any two distinct elements, both of which square to $$h$$, is zero.

Are there other algebras?
We haven't checked completely, but here is how we would start the classification. First, we note that since the algebra group is commutative, the algebra must also be commutative. The other important idea we use is that powering map by field characteristic is same in algebra and algebra group.

Suppose $$N$$ is a nilpotent associative algebra with algebra group isomorphic to direct product of Z4 and Z2. Let $$u$$ be an element of $$N$$ such that $$1 + u$$ has order four in the algebra group. Let $$v = u^2$$. We get that $$(1 + u)^2 = 1 + v$$, so $$v \ne 0$$. Also, because $$1 + u$$ has order four, $$(1 + v)^2 = 1$$ so $$v^2 = 0$$.

Also note that $$(u + v)^2 = u^2 + 2uv + v^2 = v + 0 + 0 = v$$, so $$1 + (u + v)$$ also has order four.

Let $$w$$ be an element of $$N$$ such that $$\langle 1 + w \rangle$$ is of order two and intersects $$\langle 1 + u \rangle$$ trivially. Note that $$w$$ is not equal to $$0,u,v,u+v$$, i.e., it is not in the linear span of $$u$$ and $$v$$. So, $$u,v,w$$ form a basis for $$N$$. Since $$1 + w$$ has order two, $$w^2 = 0$$, so the multiplication table so far is:

Case that $$uw$$ is in the span of $$u$$ and $$w$$
There are four possibilities for $$uw$$: $$u,w,u+w,0$$. We rule out the nonzero cases:


 * $$uw = u$$: This contradicts associativity as follows. $$(uw)w = uw = u$$ but $$u(ww) = u(0) = 0$$.
 * $$uw = w$$: This contradicts associativity as follows. $$u(u(u(uw))) = u(u(uw)) = u(uw) = uw = w$$ but $$(u^4)w = (u^2)^2w = v^2w = 0w = 0$$.
 * $$uw = u + w$$: This contradicts associativity as follows. $$(uw)w = (u + w)w = uw + w^2 = uw = u + w$$ but $$u(ww) = u(0) = 0$$.

The only case left is $$uw = 0$$. By commutativity, $$wu = 0$$. By associativity, $$vw = (u^2)w = u(uw) = u(0) = 0$$ and again by commutativity $$wv = 0$$. We thus get:

We now need to consider the product $$uv$$. Using similar reasoning to the above, we can rule out $$uv = u,v,u + v$$. We can similarly rule out $$uv = u + w, v + w, u + v + w$$. The only possibilities left are $$uv = 0$$ and $$uv = w$$. These two cases give us $$N_1$$ and $$N_2$$ respectively. For $$N_1$$, $$a = u, b = v, c = w$$ and for $$N_2$$, $$x = u, y = v, z = w$$.

Case that $$uw$$ is not in the span of $$u$$ and $$w$$
This case still needs to be explored.