What Are the Subalgebras of Biquaternions?

[EinsteinKreuz]

So upon reading the wikipedia entry about the biquaternions I noticed that this algebra has several interesting subalgebras:

1. The split-complex numbers of the form {σ = x+y(hi)| ∀(x,y)∈ ℝ} which have the norm σ⋅σ^* = (x²-y²).

2. The tessarines which can be written as {α + βj | ∀(α,β,)∈ℂ¹ & j² = -1}

3. The coquaternions whose bases form the dihedral group D4 and are define as the Span{1, i, (hj), (hk)}

But there is a 4th subalgebra that is somewhat similar to the coquaternions. And its elements can be defined as {g = a + bi + ch + dhi | ∀(a,b,c,d)∈ℝ}. Now of course i² = h² = -1 and hi² =
h²i² = (-1)² = +1.

But since I'm not sure how to add a grid for the Cayley table I'll also wrote down the other relational equations:

h⋅i = +(hi)
i⋅h = -(hi)
(hi)⋅i = -h
i⋅(hi) = +h
h⋅(hi) = -i
(hi)⋅i = +i

Using these rules it can be shown that { g | g ∈ Span[1,i,h,(hi)]} is closed under products and if we define g^* = a - bi - ch - dhi, then g⋅g^* = a²+b²+c²-d² = -ds² where ds² is the Minkowski metric.

So does this subalgebra have an official name and could it's elements be used as operators to describe the Lorentz transformation?

 

[Samy_A]

h⋅i = +(hi)
i⋅h = -(hi)
(hi)⋅i = -h
i⋅(hi) = +h
h⋅(hi) = -i
(hi)⋅i = +i

A question, as I'm not familiar with Biquaternions:
How did you get i⋅h = -(hi)?

 

[EinsteinKreuz]

A question, as I'm not familiar with Biquaternions:
How did you get i⋅h = -(hi)?

Ahhhh...good point! The article states that h⋅i = i⋅h.
Now after tinkering with my original rule I realize that if the product of h with i,j, or k is associative(like the quaternions) then i⋅h = -(hi) → (hi)² = -1.
So to generate an algebra where i⋅h = -(hi) & (hi)² = h²i² = +1, then the product of h with i,j, or k must be non-associative.