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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m (Vipul moved page Algebra group is isomorphic to subgroup of unitriangular matrix group of degree one more than logarithm of order to base of field size to [[Algebra group is isomorphic to algebra subgroup of unitriangular matrix group of degree one...)
(Statement)
 
(5 intermediate revisions by the same user not shown)
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# The order of <math>G</math> is a power of <math>q</math>.
 
# The order of <math>G</math> is a power of <math>q</math>.
# <math>G</math> must be isomorphic to a subgroup of the [[unitriangular matrix group]] <math>UT(m + 1,q)</math> where <math>m = \log_q(|G|)</math>.
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# <math>G</math> must be isomorphic to a subgroup of the [[unitriangular matrix group]] <math>UT(m + 1,q)</math> where <math>m = \log_q(|G|)</math>, 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 <math>NT(m + 1,q)</math> of strictly upper triangular matrices. Another way of saying this is that <math>G</math> is isomorphic to an [[algebra subgroup]] of <math>UT(m+1,q)</math>.
  
 
===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>.
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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> of strictly upper triangular matrices. Another way of saying this is that <math>G</math> is isomorphic to an [[algebra subgroup]] of <math>UT(n+1,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>.
 
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>.
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===Complete characterization of algebra groups===
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 +
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==
 
==Related facts==
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'''Given''': <math>G</math> is a finite <math>p</math>-group that is an algebra group over <math>\mathbb{F}_q</math>, where <math>q</math> is a power of <math>p</math>.
 
'''Given''': <math>G</math> is a finite <math>p</math>-group that is an algebra group over <math>\mathbb{F}_q</math>, where <math>q</math> is a power of <math>p</math>.
  
'''To prove''': The order of <math>G</math> is a power of <math>q</math> and <math>G</math> is isomorphic to a subgroup of <math>UT(m + 1,q)</math> where <math>m = \log_q(|G|)</math>.
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'''To prove''': The order of <math>G</math> is a power of <math>q</math> and <math>G</math> is isomorphic to a subgroup of <math>UT(m + 1,q)</math> where <math>m = \log_q(|G|)</math>. Further, subtracting 1 from all the elements of this subgroup gives a subring of <math>NT(m+1,q)</math>.
  
 
{| class="sortable" border="1"
 
{| class="sortable" border="1"
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| 2 || The order of <math>G</math> equals the order of <math>N</math>, and <math>m = \log_q(|G|)</math> is an integer that equals the dimension of <math>N</math> as a vector space over <math>\mathbb{F}_q</math>.|| || || Step (1) ||
 
| 2 || The order of <math>G</math> equals the order of <math>N</math>, and <math>m = \log_q(|G|)</math> is an integer that equals the dimension of <math>N</math> as a vector space over <math>\mathbb{F}_q</math>.|| || || Step (1) ||
 
|-
 
|-
| 3 || <math>G</math> is a subgroup of the algebra <math>N + F</math> (the unitization of <math>N</math>, with dimension <math>m + 1</math>) and acts faithfully on <math>N + F</math> by left multiplication. || || || Steps (1), (2) ||
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| 3 || <math>N + F</math> (the unitization of <math>N</math>) has dimension <math>m + 1</math>, and it acts faithfully on itself by left multiplication.  || || || Steps (1), (2) ||
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|-
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| 4 || <math>N + F</math> is isomorphic to a subring of the matrix ring <math>M(n + 1,F)</math>. Under this identification, <math>G</math>, as a multiplicative subgroup, is isomorphic to a multiplicative subgroup of <math>GL(n+1,F)</math> and <math>N</math> is a nilpotent subring of <math>M(n+1,F)</math>. || || || Step (3) ||
 
|-
 
|-
| 4 || <math>G</math> is isomorphic to a subgroup of <math>GL(m + 1,F)</math> || || || Step (3) ||
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| 5 || The image of <math>G</math> under the isomorphism of Step (4) is conjugate to a subgroup of <math>UT(m + 1,F)</math>, which is the <math>p</math>-Sylow subgroup of <math>GL(m+1,F)</math>. Moreover, subtracting 1 from all the elements of this subgroup gives a nilpotent associative subring of <math>NT(m+1,F)</math> (the strictly upper triangular matrices).  || Fact (1) || <math>G</math> is a <math>p</math>-group || Step (4) || By Step (4) and Fact (1), we know that the image of <math>G</math> in <math>GL(m + 1,F)</math> is conjugate to a subgroup of <math>UT(m+1,F)</math>, the upper triangular matrices. Under this conjugation operation, <math>N</math> goes to a new ring. Since subtracting 1 commutes with conjugation, this conjugate of <math>N</math> is the same as the ring obtained by subtracting 1 from the conjugate of <math>G</math>. Hence, it is a subring of <math>UT(m+1,F)</math>.
 
|-
 
|-
| 5 || The image of <math>G</math> under the isomorphism of Step (4) to a subgroup of <math>UT(m + 1,F)</math>, which is the <math>p</math>-Sylow subgroup of <math>GL(m+1,F)</math> || Fact (1) || <math>G</math> is a <math>p</math>-group || Step (4) ||
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| 6 || The image of <math>G</math> under the isomorphism of Step (4) is conjugate to a subgroup of <math>UT(m + 1,F)</math> with the property that subtracting 1 from all its element gives a subring of <math>NT(m + 1,F)</math>. || || || Step (5) || Follows from Step (5) and the observation that conjugation preserves the isomorphism type.
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|}
 
|}
 
|}
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===Proof for prime field===
 
===Proof for prime field===
  
 
This follows from the general proof by setting <math>q = p</math> and noting that <math>n = m</math> in that case.
 
This follows from the general proof by setting <math>q = p</math> and noting that <math>n = m</math> in that case.
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==References==
 
==References==
  
 
* {{mathoverflow|number=68101|title=p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra|details = Jack Schmidt's answer sketches a concrete version of the argument to show that [[Z8 is not an algebra group]]}}
 
* {{mathoverflow|number=68101|title=p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra|details = Jack Schmidt's answer sketches a concrete version of the argument to show that [[Z8 is not an algebra group]]}}

Latest revision as of 22:13, 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) 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