# 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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# 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>, 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>. | + | # <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>. 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>. | + | 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>. | ||

+ | |||

+ | ===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== | ==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>. | + | '''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> | + | | 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) || |

+ | |- | ||

+ | | 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) || | ||

|- | |- | ||

− | | | + | | 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>. |

|- | |- | ||

− | | | + | | 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. |

+ | |} | ||

|} | |} | ||

+ | |||

===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. | ||

+ | |||

==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

## Contents

## Statement

### Version for algebra group over prime field

Suppose is a prime number, and is a finite p-group, i.e., a group of order a power of . Suppose is a power of . Then, if is an algebra group over , the following must be true:

- The order of is a power of .
- must be isomorphic to a subgroup of the unitriangular matrix group where , 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 of strictly upper triangular matrices. Another way of saying this is that is isomorphic to an algebra subgroup of .

### Version for algebra group over prime field

Suppose is a prime number, and is a finite p-group, i.e., a group of order for some nonnegative integer . Then, if is an algebra group over , must be isomorphic to a subgroup of the unitriangular matrix group . 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 of strictly upper triangular matrices. Another way of saying this is that is isomorphic to an algebra subgroup of .

Note that if is an algebra group over any field extension of , it is also an algebra group over , so if this condition is violated, cannot be an algebra group over any for a power of .

### 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 -subgroup of a finite group is contained in a -Sylow subgroup and hence has a conjugate contained in a predetermined -Sylow subgroup.

## Proof

### Proof for finite field

**Given**: is a finite -group that is an algebra group over , where is a power of .

**To prove**: The order of is a power of and is isomorphic to a subgroup of where . Further, subtracting 1 from all the elements of this subgroup gives a subring of .

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Let be a nilpotent associative algebra over for which is the corresponding algebra group. | is an algebra group. | |||

2 | The order of equals the order of , and is an integer that equals the dimension of as a vector space over . | Step (1) | |||

3 | (the unitization of ) has dimension , and it acts faithfully on itself by left multiplication. | Steps (1), (2) | |||

4 | is isomorphic to a subring of the matrix ring . Under this identification, , as a multiplicative subgroup, is isomorphic to a multiplicative subgroup of and is a nilpotent subring of . | Step (3) | |||

5 | The image of under the isomorphism of Step (4) is conjugate to a subgroup of , which is the -Sylow subgroup of . Moreover, subtracting 1 from all the elements of this subgroup gives a nilpotent associative subring of (the strictly upper triangular matrices). | Fact (1) | is a -group | Step (4) | By Step (4) and Fact (1), we know that the image of in is conjugate to a subgroup of , the upper triangular matrices. Under this conjugation operation, goes to a new ring. Since subtracting 1 commutes with conjugation, this conjugate of is the same as the ring obtained by subtracting 1 from the conjugate of . Hence, it is a subring of . |

6 | The image of under the isomorphism of Step (4) is conjugate to a subgroup of with the property that subtracting 1 from all its element gives a subring of . | 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 and noting that in that case.

## References

- MathOverflow question: p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra: Jack Schmidt's answer sketches a concrete version of the argument to show that Z8 is not an algebra group