Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size

Version for algebra group over prime field
Suppose $$p$$ is a prime number, and $$G$$ is a finite p-group, i.e., a group of order a power of $$p$$. Suppose $$q$$ is a power of $$p$$. Then, if $$G$$ is an algebra group over $$\mathbb{F}_q$$, the following must be true:


 * 1) The order of $$G$$ is a power of $$q$$.
 * 2) $$G$$ must be isomorphic to a subgroup of the unitriangular matrix group $$UT(m + 1,q)$$ where $$m = \log_q(|G|)$$, and the image of this isomorphism must have the further property that subtracting 1 from all its elements gives a subring of the nilpotent associative ring $$NT(m + 1,q)$$ of strictly upper triangular matrices. Another way of saying this is that $$G$$ is isomorphic to an algebra subgroup of $$UT(m+1,q)$$.

Version for algebra group over prime field
Suppose $$p$$ is a prime number, and $$G$$ is a finite p-group, i.e., a group of order $$p^n$$ for some nonnegative integer $$n$$. Then, if $$G$$ is an algebra group over $$\mathbb{F}_p$$, $$G$$ must be isomorphic to a subgroup of the unitriangular matrix group $$UT(n + 1,p)$$. Further, the image of this isomorphism must have the further property that subtracting 1 from all its elements gives a subring of the nilpotent associative ring $$NT(m + 1,q)$$ of strictly upper triangular matrices. Another way of saying this is that $$G$$ is isomorphic to an algebra subgroup of $$UT(n+1,p)$$.

Note that if $$G$$ is an algebra group over any field extension of $$\mathbb{F}_p$$, it is also an algebra group over $$\mathbb{F}_p$$, so if this condition is violated, $$G$$ cannot be an algebra group over any $$\mathbb{F}_q$$ for $$q$$ a power of $$p$$.

Complete characterization of algebra groups
Note that it's obviously true that any algebra subgroup of a unitriangular matrix group is an algebra group. So the above gives a complete (necessary and sufficient) characterization of algebra groups.

Applications

 * Z8 is not an algebra group
 * Cyclic group of prime-square order is not an algebra group for odd prime

Facts used

 * 1) uses::Sylow's theorem, specifically uses::Sylow implies order-dominating which says that any $$p$$-subgroup of a finite group is contained in a $$p$$-Sylow subgroup and hence has a conjugate contained in a predetermined $$p$$-Sylow subgroup.

Proof for finite field
Given: $$G$$ is a finite $$p$$-group that is an algebra group over $$\mathbb{F}_q$$, where $$q$$ is a power of $$p$$.

To prove: The order of $$G$$ is a power of $$q$$ and $$G$$ is isomorphic to a subgroup of $$UT(m + 1,q)$$ where $$m = \log_q(|G|)$$. Further, subtracting 1 from all the elements of this subgroup gives a subring of $$NT(m+1,q)$$.


 * }

Proof for prime field
This follows from the general proof by setting $$q = p$$ and noting that $$n = m$$ in that case.