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

## 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