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"

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(Version for algebra group over prime field)
(Statement)
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===Version for algebra group over prime field===
 
===Version for algebra group over prime field===
  
Suppose <math>p</math> is a [[prime number]], and <math>G</math> is a [[finite p-group]], i.e., a group of order <math>p^n</math> for some nonnegative integer <math>n</math>. Then, if <math>G</math> is an [[algebra group]][[fact about::algebra group;2| ]] over <math>\mathbb{F}_p</math>, <math>G</math> must be isomorphic to a subgroup of the [[unitriangular matrix group]] <math>UT(n + 1,p)</math>.
+
Suppose <math>p</math> is a [[prime number]], and <math>G</math> is a [[finite p-group]], i.e., a group of order <math>p^n</math> for some nonnegative integer <math>n</math>. Then, if <math>G</math> is an [[algebra group]][[fact about::algebra group;2| ]] over <math>\mathbb{F}_p</math>, <math>G</math> must be isomorphic to a subgroup of the [[unitriangular matrix group]] <math>UT(n + 1,p)</math>. 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 <math>NT(m + 1,q)</math>.
  
 
Note that if <math>G</math> is an algebra group over any field extension of <math>\mathbb{F}_p</math>, it is also an algebra group over <math>\mathbb{F}_p</math>, so if this condition is violated, <math>G</math> cannot be an algebra group over any <math>\mathbb{F}_q</math> for <math>q</math> a power of <math>p</math>.
 
Note that if <math>G</math> is an algebra group over any field extension of <math>\mathbb{F}_p</math>, it is also an algebra group over <math>\mathbb{F}_p</math>, so if this condition is violated, <math>G</math> cannot be an algebra group over any <math>\mathbb{F}_q</math> for <math>q</math> a power of <math>p</math>.

Revision as of 21:52, 16 August 2012

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).

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).

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.

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|).

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 G is a subgroup of the algebra N + F (the unitization of N, with dimension m + 1) and acts faithfully on N + F by left multiplication. Steps (1), (2)
4 G is isomorphic to a subgroup of GL(m + 1,F) Step (3)
5 The image of G under the isomorphism of Step (4) to a subgroup of UT(m + 1,F), which is the p-Sylow subgroup of GL(m+1,F) Fact (1) G is a p-group Step (4)

Proof for prime field

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

References