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

Statement

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:

The order of $G$ is a power of $q$ .
$G$ 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.

Related facts

Applications

Facts used

Sylow's theorem, specifically 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

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 $NT(m+1,q)$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $N$ be a nilpotent associative algebra over $\mathbb{F}_q$ for which $G$ is the corresponding algebra group.		$G$ is an algebra group.
2	The order of $G$ equals the order of $N$ , and $m = \log_q(\|G\|)$ is an integer that equals the dimension of $N$ as a vector space over $\mathbb{F}_q$ .			Step (1)
3	$N + F$ (the unitization of $N$ ) has dimension $m + 1$ , and it acts faithfully on itself by left multiplication.			Steps (1), (2)
4	$N + F$ is isomorphic to a subring of the matrix ring $M(n + 1,F)$ . Under this identification, $G$ , as a multiplicative subgroup, is isomorphic to a multiplicative subgroup of $GL(n+1,F)$ and $N$ is a nilpotent subring of $M(n+1,F)$ .			Step (3)
5	The image of $G$ under the isomorphism of Step (4) is conjugate to a subgroup of $UT(m + 1,F)$ , which is the $p$ -Sylow subgroup of $GL(m+1,F)$ . Moreover, subtracting 1 from all the elements of this subgroup gives a nilpotent associative subring of $NT(m+1,F)$ (the strictly upper triangular matrices).	Fact (1)	$G$ is a $p$ -group	Step (4)	By Step (4) and Fact (1), we know that the image of $G$ in $GL(m + 1,F)$ is conjugate to a subgroup of $UT(m+1,F)$ , the upper triangular matrices. Under this conjugation operation, $N$ goes to a new ring. Since subtracting 1 commutes with conjugation, this conjugate of $N$ is the same as the ring obtained by subtracting 1 from the conjugate of $G$ . Hence, it is a subring of $UT(m+1,F)$ .
6	The image of $G$ under the isomorphism of Step (4) is conjugate to a subgroup of $UT(m + 1,F)$ with the property that subtracting 1 from all its element gives a subring of $NT(m + 1,F)$ .			Step (5)	Follows from Step (5) and the observation that conjugation preserves the isomorphism type.