Classification of finite p-groups of normal rank one
This article gives a classification statement for certain kinds of groups of prime power order, subject to additional constraints.
View other such statements
- If is odd, is itself cyclic.
- If , is either cyclic, or it has a cyclic maximal subgroup, with the quotient acting by multiplication by either (dihedral group and generalized quaternion group) or , where .
- Classification of finite p-groups of characteristic rank one
- Classification of finite p-groups of rank one
- Classification of finite p-groups of characteristic rank one: If is a finite -group of characteristic rank one, is a central product of subgroups where is trivial or extraspecial, and is cyclic for odd , and for , is either cyclic or it has a cyclic maximal subgroup, with the quotient acting by multiplication by either (dihedral group and generalized quaternion group) or , where .
- Central factor implies transitively normal
- Extraspecial and normal rank one implies quaternion group
Given: A finite -group of normal rank one.
To prove: For odd, is cyclic. For , is a dihedral group, generalized quaternion group, or is obtained as a semidirect product of a cyclic maximal subgroup with a subgroup acting on it via multiplication by where the order of is .
Proof': Since has normal rank one, it also has characteristic rank one, and thus is a finite -group of characteristic rank one. By fact (1), we can write with the conditions specified in fact (1). Note also that cannot be trivial, otherwise we would have Abelian normal non-cyclic subgroups such as the product of nontrivial cyclic normal subgroups of both.
Now, since is a central factor of , fact (2) states that any normal subgroup of is normal in . Thus, any Abelian normal subgroup of is an Abelian normal subgroup of , and is thus cyclic. Thus, itself has normal rank one. Applying fact (3) tells us that either is trivial or and is the quaternion group.
For odd, is trivial, and . Since is cyclic, is cyclic and we are done.
For , we make two cases:
- is trivial: In this case, , and we observe that the possibilities for are precisely the possibilities we need to prove are permisslbe for .
- is the quaternion group: If has order two, being nontrivial, must equal . In this case, is the quaternion group. In case has order bigger than two, the structural possibilities for show that there must exist a cyclic normal subgroup of of order four. Let be a cyclic normal subgroup of order four in . Since and commute element-wise, so do and so is an Abelian normal subgroup of of exponent four. Further, since is nontrivial and central, , so and . Thus, is noncyclic, leading to a contradiction. Thus, we conclude that the case of a quaternion group cannot hold here.