Algebra group structures for direct product of Z4 and Z2

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This article gives specific information, namely, algebra group structures, about a particular group, namely: direct product of Z4 and Z2.
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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.

Summary

Algebra Smallest n for which it can be embedded in UT(n,2) Maximum of element nilpotencies Nilpotency index of whole algebra
direct product of algebras that give cyclic group:Z4 and cyclic group:Z2 (we call this N_1) 4 3 3
has nonzero products of length three (we call this N_2) 4 4 4
all products of length three are zero, but not a direct product (we call this N_3) 4 3 3

Algebra that splits as a direct product, all products of length three are zero

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).

a b c
a b 0 0
b 0 0 0
c 0 0 0

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}

Algebra with nonzero products of length three

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).

x y z
x y z 0
y z 0 0
z 0 0 0

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}

Algebra with zero products of length three, but not a direct product

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).

g h k
g h 0 h
h 0 0 0
k h 0 0

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:

u v w
u v  ?  ?
v  ? 0  ?
w  ?  ? 0

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:

u v w
u v  ? 0
v  ? 0 0
w 0 0 0

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.