# Difference between revisions of "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:

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.

## Facts used

1. 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.

|}

### Proof for prime field

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