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 ![]() ![]() ![]() |
![]() |
|||
2 | The order of ![]() ![]() ![]() ![]() ![]() |
Step (1) | |||
3 | ![]() ![]() ![]() |
Steps (1), (2) | |||
4 | ![]() ![]() ![]() ![]() ![]() ![]() |
Step (3) | |||
5 | The image of ![]() ![]() ![]() ![]() ![]() |
Fact (1) | ![]() ![]() |
Step (4) | By Step (4) and Fact (1), we know that the image of ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
6 | The image 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