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

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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.

Related facts

Applications

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.

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Proof for prime field

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

References